PS

DEFINITION OF A SEQUENCEA succession of numbers a1, a2, a3..., an, ... formed, according to some definite rule, is called a sequence.

ARITHMETIC PROGRESSION (A.P.)A sequence of numbers {an} is called an arithmetic progression, if there is a number d, such thatd = an–an–1 for all n. d is called the common difference (C.D.) of the A.P.(i) Useful FormulaeIf a = first term, d = common difference and n is the number of terms, then(a) nth term is denoted by tn and is given by

tn = a + (n – 1) d.(b) Sum of first n terms is denoted by Sn and is given by

nnS [2a (n 1)d]2

or ( )2

nnS a l , where l = last term in the series i.e., l = tn = a + (n – 1) d.

(c) If terms are given in A.P., and their sum is known, then the terms must be picked up in followingway

For three terms in A.P., we choose them as (a – d), a, (a + d) For four terms in A.P. , we choose them as (a – 3d), (a – d), (a + d), (a + 3d) For five terms in A.P., we choose them as (a – 2d), (a – d), a, (a + d), (a + 2d) etc.(ii) Useful Properties If tn = an + b, then the series so formed is an A.P. If Sn = an2 + bn then series so formed is an A.P. If every term of an A.P. is increased or decreased by some quantity, the resulting terms will also

be in A.P. If every term of an A.P. is multiplied or divided by some non-zero quantity, the resulting terms will

also be in A.P. In an A.P. the sum of terms equidistant from the beginning and end is constant and equal to sum

of first and last terms. Sum and difference of corresponding terms of two A.P.’s will form an A.P.

If terms a1, a2, ..., an, an+1, ..., a2n+1 are in A.P., then sum of these terms will be equal to(2n + 1)an+1.

If terms a1, a2, ..., a2n–1, a2n are in A.P. The sum of these terms will be equal to

(2n) n n 1a a

2

.

Illustration 1:The mth term of an A.P. is n and its nth term is m. Prove that its pth term is m + n – p. Also showthat its (m + n) th term is zero.

Solution:Given Tm = a + (m – 1) d = n and Tn = a + (n – 1) d = mSolving we get, d = –1 and a = m + n – 1 Tp = a + (p –1)d = m + n – 1 + (p – 1) (–1) = m + n – pNow, Tm + n = a + (m + n – 1)d = (m + n –1) + (m + n–1) (–1) = 0.

PROGRESSION & SERIES

2 PROGRESSION & SERIES

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Illustration 2:

Find the number of terms in the series 20, 1 219 , 18 ,3 3 ..... of which the sum is 300. Explain the

double answer.Solution:

Clearly here a = 20, 2d3

and Sn = 300.

n 22 20 (n 1) 3002 3

. Simplifying, n2 – 61n + 900 = 0 n = 25 or 36.

Since common difference is negative and S25 = S36 = 300, it shows that the sum of the eleven termsi.e., T26, T27 , ....., T36 is zero.

Illustration3:

In an A. P., if the pth term is q1

and the qth term is p1

, prove that the sum of the first pq terms must be

21

(pq + 1).

Solution:

p1T a p 1 dq

q1&T a q 1 dp

Solving Tp & Tq, are get

1a dpq

pqpq pq 1S 2a pq 1 d2 2

Illustration 4:If the sum of n terms of an A. P. is (pn + qn2), where p and q are constants, find the common difference.

Solution:

nnS 2a n 1 d2

2d dn a n2 2

on comparing Sn with given sum

d da p and q2 2

a p q & d 2q

3PROGRESSION & SERIES

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Illustration 5:

If a + b + c ¹ 0 and acb

, bac

, cba

are in A. P., prove that a1

, b1

, c1

are also in A. P.A. P.

Solution:

b c c a a b1, 1, 1a b c

also in A.P..

a b c c a b a b c, ,a b c

are in A.P..

1 1 1, ,a b c

are in A.P..

GEOMETRIC PROGRESSION (G.P.)

A sequence of the numbers {an}, in which 1 0a , is called a geometric progression, if there is a number

0r such that n

n 1

a ra

for all n then r is called the common ratio (C.R.) of the G.P..

(i) Useful FormulaeIf a = first term, r = common ratio and n is the number of terms, then(a) nth term, denoted by tn , is given by tn = arn–1

(b) Sum of first n terms denoted by Sn is given by

n

na(1 r )S

1 r

or na(r 1)

r 1

In case r = 1, Sn = na.

(c) Sum of infinite terms (S )

aS (for | r | 1& r 0)1 r

(d) If terms are given in G.P. and their product is known, then the terms must be picked up in thefollowing way.

For three terms in G.P., we choose them as a , a, arr

For four terms in G.P., we choose them as 3

3

a a, , ar, arr r

For five terms in G.P., we choose them as 2

2

a a, , a,ar, arr r etc.

(ii) Useful Properties(a) The product of the terms equidistant from the beginning and end is constant, and it is equal to the

product of the first and the last term.(b) If every term of a G.P. is multiplied or divided by the some non-zero quantity, the resulting progression

is a G.P.


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(c) If a1, a2, a3 ... and b1, b2, b3, ... be two G.P.’s of common ratio r1 and r2 respectively, then a1b1 , a2b2

... and 31 2

1 2 3

aa a, ,b b b

... will also form a G.P. Common ratio will be r1r2 and 1

2

rr

respectively..

(d) If a1, a2, a3, ... be a G.P. of positive terms, then loga1, loga2, loga3, ... will be an A.P. and conversely.

Illustration 6:The first term of an infinite G..P is 1 and any term is equal to the sum of all the succeeding terms. findthe series.

Solution:Given that Tp = (Tp+1 + Tp+2 + ..... ) or, arp–1 = arp + arp+1 + arp+2 + ...

rp–1 = pr

1 r [sum of an infinite G.P.]

1 – r = r r = 12 . Hence the series is

1 1 11, , , , ...2 4 8

.

Illustration 7:If the first and the nth terms of a G. P. are a and b, respectively, and if P is the product of first n terms, prove thatP2 = (ab)n.Solution:

b = arn–1 ......... (i)

2 n 1p a ar ar ............. ar

n(n 1)n 1 2 ............. n 1 n 2a r a r

n n

2 n 1 n 12 2p a r a.ar

n2ab

or 2 np ab

Illustration 8:If a G. P. the first term is 7, the last term 448, and the sum 889; find the common ratio.Solution:

a = 7, l = arn–1 = 448

n

n

a r 1S 889

r 1

Here n 1

n

r. a r a 448r aSr 1 r 1

r 2 & a 7

HARMONIC PROGRESSION (H.P.)A sequence is said to be in harmonic progression, if and only if the reciprocal of its terms form anarithmetic progression.For example


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1 1 1, ,2 4 6

... form an H.P., because 2, 4, 6, ... are in A.P..

(a) If a, b are first two terms of an H.P. then

n1t

1 1 1(n 1)a b a

(b) There is no formula for sum of n terms of an H.P.(c) If terms are given in H.P. then the terms could be picked up in the following way

l For three terms in H.P, we choose them as

1 1 1, ,a d a a d

l For four terms in H.P, we choose them as 1 1 1 1, , ,

a 3d a d a d a 3d

l For five terms in H.P, we choose them as 1 1 1 1 1, , , ,

a 2d a d a a d a 2d ii) Useful properties:

If every term of a H.P. is multiplied or divided by some non zero fixed quantity, theresulting progression is a H.P.

Illustration 9:If a1, a2, a3, ... an are in harmonic progression, prove thata1a2 + a2a3 + ... + an–1 an = (n – 1) a1an .

Solution:Since a1, a2 , ... , an are in H.P.,

1 2 3 n

1 1 1 1, , ,...,a a a a are in A.P. having common difference d (say) .

2 1 3 2 n n 1

1 1 1 1 1 1d, d, ... da a a a a a

or a1 – a2 = d(a1a2), a2 – a3 = d (a2a3), ... , (an–1 – an) = d(an–1 an )Adding the above relations, we geta1 – an = d (a1a2 + a2a3 +... + an–1 an) ... (1)

Now n 1

1 1 (n 1)da a

n 1

1 1 (n 1)da a

or (a1 – an) = (n – 1) d an a1 ... (2)Putting the value of a1 – an from (2) in (1), we get(n – 1) an a1d = d (a1a2 + a2a3 + ... + an–1 an) (n – 1) ana1 = a1a2 + a2a3 + ... + an–1 an.


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Illustration 10:

Find H. P. whose 3rd and 14th terms are respectively 76

and 31

.

Solution:Let a & d are first term & common difference of A.P. which is reciprocal of given H.P.

3 147t a 2d & t 3 a 13d6

5 1a & d6 6

There A.P. is 5 7 8 9,1, , , ..........6 6 6 6 &

H.P. is 6 6 6 6, 1, , , ..........5 7 8 9

Illustration 11:If the pth, qth and rth terms of a H. P. are a, b, c respectively, prove that

arq

+ bpr

+ cqp

= 0

Solution:Let Tp, Tq, Tr are pth, qth & rth term of a H.P.

p q

1 1 1 1A p 1 d, A q 1 dT a T b

and r

1 1 A r 1 dT C

Hence q r r p p q A p 1 d) q r A q 1 d (r p A r 1 d p q 0a b c

INSERTION OF MEANS BETWEEN TWO NUMBERSLet a and b be two given numbers.(i) Arithmetic Meansl If three terms are in A.P. then the middle term is called the arithmetic mean (A.M.)

between the other two i.e. if a, b, c are in A.P. then a cb

2

is the A.M. of a and b.

l If a, A1, A2 , ... An, b are in A.P., then A1, A2, ... An are called n A.M.’s between a and b.

If d is the common difference, then b = a + (n + 2 – 1) d d = b an 1

Ai = a + id = a + i b a a(n 1 i) ib ,n 1 n 1

i = 1, 2, 3, ..., n

Note: The sum of n-A. Ms, i.e., AA1 + A2 + ... + An = n (a b)2


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(ii) Geometric meansl If three terms are in G.P. then the middle term is called the geometric mean (G.M.)between the two. So if a, b, c are in G.P. then b ac or b ac corresponding to a &c both are positive or negative respectively.l If a, G1, G2 ... Gn, b are in G.P., then G1, G2 ... Gn are called n G.M.s between a and b. If

r is the common ratio, then b = a.rn+1 r = 1

(n 1)ba

Gi = ari =

in 1 i in 1n 1 n 1ba a .b ,

a

i = 1, 2, ..., n

Note: The product of n-G. Ms i.e., G1 G2 ... Gn = nab

(iii) Harmonic mean:l If a & b are two non-zero numbers, then the harmonic mean of a & b is a number H

such that a, H, b are in H.P. & 1 1 1 1 2abor HH 2 a b a b

l If a, H1, H2... Hn, b are in H.P., then H1, H2 ... Hn are called n H.M. s between a and b. Ifd is the common difference of the corresponding A.P., then

1 1 a b(n 2 1)d db a ab(n 1)

i

1 1 1 a bid iH a a ab(n 1)

,Hi ab(n 1) , i 1, 2, 3, ..., n

b(n i 1) ia

(iv) Term tn+1 is the arithmetic, geometric or harmonic mean of t1 & t2n+1 accordingas the terms t1, tn+1 t2n+1 are in A.P., G.P. or H.P. respectively.

Illustration 12:If A1, A2; G1 , G2 and H1 , H2 be two A.M.s, G.M.s and H.M.s between two quantities ‘a’ and ‘b’ thenshow that A1H2 = A2 H1 = G1G2 = ab

Solution:a, A1 , A2, b be are in A.P. ... (1)a, H1 , H2 , b are in H.P.

1 2

1 1 1 1, , ,a H H b are in A.P..

Multiply by ab.

1 2

ab abb, , , a are in A.P.H H

take in reverse order or 2 1

ab aba, , ,bH H are in A.P.. ... (2)


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Compare (1) and (2)

1 22 1

ab abA and AH H

A1H2 = A2H1 = ab = G1G2

Illustration 13:Between 2 & 100, 13 means are inserted then find the 9th mean if means arei) arithmetic ii) geometric iii) harmonicSolution:i. a = 2, b = 100, n = 13

9b aA a 9d a 9n 1

= 65ii. a = 2, b = 100, n = 13

9149

9bG ar aa

9

142 50

iii) a = 2, b = 100, n = 13 reciprocal of harmonic is A.P. where 1 151 1t & t2 100

10 19

1 t t 9dH

1 792 200

Illustration 14:

If H.M. & A.M. of two numbers are 3 & 4 respectively, find the numbers.

Solution:

Let numbers are a & b

2ab a b3 & 4a b 2

ab 12 and a b 8

solving there we get a = 6, b = 2 or a = 2, b = 6

Illustration 15:

The sum of the two numbers is 6 times their geometric mean. Show that the number are in the ratio

3 + 2 2 : 3 – 2 2 .

Solution:

given a b 6 ab

a b 6b a

aHere tb

21 a 3 2 2t 6 3 2 2t b 3 2 2


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Illustration 16:

If b is the harmonic mean between a and c, prove that ab1

+ cb1

= a1

+ c1

.

Solution:

2ac 1 1 2 2 1 1b2ac 2aca c b a b c a ca ca c a c

ARITHMETICO-GEOMETRIC SERIES

A series whose each term is formed, by multiplying corresponding terms of an A.P. and a G.P., is calledan Arithmetic-geometric series.

e.g. 1 + 2x + 3x2 + 4x3 + ..... ; a + (a + d) r + (a + 2d)r2 + .....

(i) Summation of n terms of an Arithmetic-Geometric Series

Let Sn = a + (a + d) r + (a + 2d)r2 + ... + [a + (n – 1)d] rn–1, d 0 , r 1

Multiply by ‘r’ and rewrite the series in the following way

rSn = ar + (a + d)r2 + (a + 2d)r3 + ... + [a + (n – 2)d]rn–1 + [a + (n – 1)d ]rn

on subtraction,

Sn (1 – r) = a + d(r + r2 + ... + rn–1) – [a + (n– 1)d]rn

or,n 1

nn

dr(1 r )S (1 r) a [a (n 1)d].r1 r

or, n 1

nn 2

a dr(1 r ) [a (n 1)d]S .r1 r (1 r) 1 r

(ii) Summation of Infinite Series

If |r| < 1, then (n –1)rn, rn–10, as n . Thus S = S = 2

a dr1 r (1 r)

Illustration 17:

Find the sum of infinity of the series 1 + 31.2

+ 231.3

+ 331.4

+ ....... .

Solution:

2 3

1 1 1S 1 2. 3. 4. .........3 3 3

2 3

1 1 1 1S 2. 3. .............3 3 3 3


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2 3

2 1 1 1S 1 .........upto in inf inite3 3 3 3

1/ 3 311 213

9S4

Illustration 18:

Find sum to n terms of the series, 1 + 4x + 7x2 + 10x3 + .... when | x | < 1.

Solution:

Tn = (3n – 2) xn–1

Sn = 1 + 4x + 7x2 + 10x3 + ..........+ (3n–2) xn–1 ..................(1)

x Sn = x + 4x2 + 7x3 + ........+ (3n–5) xn–1 + (3n–2)xn ...................(2)

On subtracting (2) from (1)

(1 – x) Sn = 1 + (3x + 3x2 + 3x3 + ........... upto (n – 1) term) – (3n–2) xn

(1 – x)Sn =

n 1

n1 x

1 3x 3n 2 x1 x

n 1 nn

1 3xS 1 1 x 3n 2 x1 x 1 x

SUM OF MISCELLANEOUS SERIES

(i) Difference Method

Suppose a1, a2,a3, ............ is a sequence such that the sequence a2 – a1, a3 – a2, ........... iseither an A.P. or G.P. The nth term ‘an’ of this sequence is obtained as follows.

S = a1 + a2 + a3 + .......... + an–1 + an

S = a1 + a2 + .......... + an–2 + an–1 + an

n 1 2 1 3 2 n n 1a a a a a a ......... a a

Since the terms within the brackets are either in an A.P. or in a G.P. we can find the value of an,

the nth term. We can now find the sum of the n terms of the sequence as n

kk 1

S a

(ii) Vn – Vn–1 Method

Let T1, T2, T3 , ... be the terms of a sequence. If there exists a sequence V1, V2, V3 ... satisfyingTk = Vk – Vk–1, k 1,

then n n

n k k k 1 n 0k 1 k 1

S T (V V ) V V

.


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Illustration 19 :

Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + ... .

Solution:

Clearly here the differences between the successive terms are

7 – 3, 14 – 7, 24 – 14, ... i.e., 4, 7, 10, ... which are in A.P.

Tn = an2 + bn + c

Thus we have 3 = a + b + c, 7 = 4a + 2b + c and 14 = 9a + 3b + c

Solving we get, 3 1a ,b ,c 22 2

. Hence 2n

1T (3n n 4)2

Sn = 21 [3 n n 4n]

2 21 n(n 1) (2n 1) n(n 1) n3 4n (n n 4)

2 6 2 2

Illustration 20 :

Find the sum of n terms of the series 3 + 8 + 22 + 72 + 266 + 1036 + .....

Solution:

1st difference 5, 14, 50, 194, 770, ...

2nd difference 9, 36, 144, 576, .....

They are in G.P. whose nth term is arn–1 = a4n–1

Tn of the given series will be of the form

Tn = a4n–1 + bn + c

T1 = a + b + c = 3

T2 = 4a + 2b + c = 8

T3 = 16a + 3b + c = 22. Solving we have a = 1, b = 2, c = 0.

Tn = 4n–1 + 2n

Sn = n 1 n14 2 n (4 1) n(n 1)3

.

INEQUALITIES

(i) A.M. G.M. H.M.

Let a1, a2, ..........., an be n positive real numbers, then we define their arithmetic mean

(A), geometric mean (G) and harmonic mean (H) as 1 2 na a ......... aAn

,


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1/ n1 2 n

1 2 3 n

nG a a .........a and H1 1 1 1.......a a a a

It can be shown that A G H. Moreover equality holds at either place if and only if

a1 = a2 = ............... = an

(ii) Weighted Means

Let a1, a2, ..........., an be n positive real numbers and w1, w2, .........., wn be n positive rational numbers.Then we define weighted Arithmetic mean (A*), weighted Geometric mean (G*) and weighted harmonicmean (H*) as

A* = 1 2 n 1 2 n

1w w w w w ...w1 1 2 2 n n

1 2 n1 2 n

a w a w ... a w , G* (a .a ...a )w w ... w

1 2 n

1 2 n

1 2 n

w w ... wand H* w w w...a a a

.

A* ³ G* ³ H* More over equality holds at either place if & only if a1 = a2i ............=.an

(iii) Cauchy’s Schwartz Inequality:

If a1, a2, .........an and b1, b2,.........., bn are 2n real numbers, then

(a1b1 + a2b2 +.............+ an bn)2 (a1

2 + a22 +............+ an

2) (b12 + b2

2 +..........+ b2n) with the

equality holding if and only ifn

n

2

2

1

1

ba..........

ba

ba

.

Illustration 21:

Prove that a b

b aa b a .b , a,b N;a b.2

Solution:

Let us consider b quantities each equal to a and a quantities each equal to b. Then sinceA.M. > G.M.

(a a a ...b times) (b b b ...a times) [(a.a.a...b times)a b

(b.b.b. ... a times)]1/(a+b)

b a 1/(a b)ab ab (a b )

a b


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b a 1/(a b)2ab (a b )

a b

Now a b 2ab

2 a b

(A.M. > H.M.)

a b

b aa b a .b2

.

ARITHMETIC MEAN OF mth POWER

Let a1, a2 ... , an be n positive real numbers and let m be a real number, then

mm m m1 2 n 1 2 na a ... a a a ... a , if m R [0,1].

n n

However if m (0, 1), then mm m m

1 2 n 1 2 na a ... a a a ... an n

Obviously if mm m m

1 2 n 1 2 na a ... a a a ... am {0,1}, thenn n

Illustration 22:

Prove that a4+ b4 + c4 abc (a + b + c), [a, b, c > 0]

Solution:

Using mth power inequality, we get

44 4 4a b c a b c3 3

3a b c a b c3 3

1/ 3 3a b c [(abc) ]

3

( A.M G.M).M)

or4 4 4a b c a b c abc

3 3

a4 + b4 + c4 abc (a + b + c).


RAMKI

Illustration 23:

Prove that s s s 9

s a s b s c 2

, if s = a + b + c, [a, b, c > 0]

Solution:

We have to prove that 1 1 1 9

b c c a a b 2(a b c)

for the proof , using mth power theorem of inequality, we get

11 1 1(a b) (b c) (c a) a b b c c a3 3

or,1 1 1 9

b c c a a b 2(a b c)

Aliter :

A.M. H.M.

(a b) (b c) (c a) 31 1 13

a b b c c a

1 1 1 9

a b b c c a 2(a b c)

KEY POINTSA.P.

If a = first term, d = common difference and n is the number of terms, then

nth term is denoted by tn and is given by

tn = a + (n – 1) d.

Sum of first n terms is denoted by Sn and is given by

nnS [2a (n 1)d]2

or ( )2

nnS a l , where l = last term in the series i.e., l = tn = a + (n – 1) d.

Arithmetic mean A of any two numbers a and b

a bA2

.

Sum of first n natural numbers ( n )

n(n 1)n2

, where n N .


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Sum of squares of first n natural numbers ( 2n )

2 ( 1) (2 1)6

n n nn

Sum of cubes of first n natural numbers 3( n )

23 n(n 1)n

2

G.P.

If a = first term, r = common ratio and n is the number of terms, then

nth term, denoted by tn , is given by tn = arn–1

Sum of first n terms denoted by Sn is given by

n

na(1 r )S

1 r

or na(r 1)

r 1

In case r = 1, Sn = na.

Sum of infinite terms (S )

aS (for | r | 1& r 0)1 r

H.P.

If a, b are first two terms of an H.P. then

n1t

1 1 1(n 1)a b a

There is no formula for sum of n terms of an H.P.

MEANS

If three terms are in A.P. then the middle term is called the arithmetic mean (A.M.)

between the other two i.e. if a, b, c are in A.P. then a cb

2

is the A.M. of a and b.

If a, A1, A2 , ... An, b are in A.P., then A1, A2, ... An are called n A.M.’s between a and b.

If d is the common difference, then b = a + (n + 2 – 1) d d = b an 1

Ai = a + id = a + i b a a(n 1 i) ib ,n 1 n 1

i = 1, 2, 3, ..., n


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If three terms are in G.P. then the middle term is called the geometric mean (G.M.)between the two. So if a, b, c are in G.P. then b ac or b ac corresponding to a & c both arepositive or negative respectively.

If a, G1, G2 ... Gn, b are in G.P., then G1, G2 ... Gn are called n G.M.s between a and b. If r is the common

ratio, then b = a.rn+1 r = 1

(n 1)ba

Gi = ari =

in 1 i in 1n 1 n 1ba a .b ,

a

i = 1, 2, ..., n

If a & b are two non-zero numbers, then the harmonic mean of a & b is a number H

such that a, H, b are in H.P. & 1 1 1 1 2abor HH 2 a b a b

If a, H1, H2... Hn, b are in H.P., then H1, H2 ... Hn are called n H.M. s between a and b. If dis the common difference of the corresponding A.P., then

1 1 a b(n 2 1)d db a ab(n 1)

i

1 1 1 a bid iH a a ab(n 1)

; Hi

ab(n 1) , i 1, 2, 3, ..., nb(n i 1) ia

CONCEPTUAL QUESTIONSSingle Answer Type Questions (A.P., G.P., & H.P.)1. The p th term of an A.P. is a and q th term is b. then the sum of its’ (p+ q) terms is

A)

2p q a ba b

p q

B)

2

p q a ba bp q

C) 2p q p qa b

a b

D) 2p q p qa b

a b

2. If the sum of m terms of an arithmetical progression is equal to the sum of either the

next n terms or the next p terms, then

n m p mn m p m is

A) np B)

pn C) np D)

pm


RAMKI

3: If Sn denotes the sum to n terms of a G.P. whose first term and common ratio are a andr(r 1) respectively, then S1 + S2 + S3 + . . . + Sn is

A) 1na

r B)

2

11 1

nar rnar r

C) 1arn

D) 2

11 1

n

n

r rar r

4. If |x| < 1 and |y| < 1 then (x + y) +(x2 + xy +y2) + (x3 +x2y+xy2 + y3 )+...... is

A) 1 1x y xy

x y B) 1 1

x y xyx y

C) 1 1x y xy

x y D) 1 1

x y xyx y

(A.M., G.M., H.M.)

5. If a, b and c be in G.P. and x, y be the arithmetic means between a, b and b, c respectively

then a cx y is

A) 2 B) 1C) 3 D) 4

6. In a centre test, there are p questions, in this 2p–r students give wrong answers to at least r questions

1 r p . If total number of wrong answers given is 2047, then the value of p is

A) 14 B) 13 C) 12 D) 11

(INEQUALITY)

7. Let a, b, c be three distinct positive real numbers in G.P., then a2 + 2bc – 3ac is

A) >0 B) <0 C) =0 D) cann’t be found out

8 If a, b, x, y are positive natural numbers such that 1 1 1x y then

x ya bx y

is

A) ab B) ab C) =ab D) cann’t be found out

9. A sequence a1, a2, a3, . . ., an of real numbers is such that a1 = 0, |a2| = |a1 + 1|, |a3| = |a2 + 1|, . . . , |an | = |an – 1 + 1|.then the arithmetic mean of a1, a2, ... , an cannot be less than

A)12 B)

12

C) 2 D) -210. The maximum value of a2 b3 c4 subject to a + b + c = 18 is

A) 2 3 44 6 8 B) 3 2 44 6 8C) 4 2 34 6 8 D) 2 4 34 6 8


RAMKI

More than One Answer (A.P., G..P., H.P.)

11. If the 1st and the 2 1 thn term of an A.P., G.P., and H.P. are equal and their nth terms are a,b

and c respectively , then

A) a b c B) a b c

C) a c b D) 2 0ac b ( Summation of seriese)

12 For a+ve integer n let 1 1 1 11 .............2 3 4 2 1na n

then

A) 100 100a B) 100 100a

C) 200 100a D) 200 100a C

Statement type Questions

Each question contains STATEMENT – 1 (Assertion) and STATEMENT – 2 (Reason). Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1(B) Statement-1 is True, Statement-2 is True; Statement-2 NOT a correct explanation for Statement-1.(C) Statement-1 is True, Statement-2 is False(D) Statement -1 is False, statement-2 is True

13. Statement - I : For positive real numbers x, y , z x y z x y z2 2 2

x y zx y z x y zx y zx y z 3

Statement - II : Weighted AM weighted GM

14 Statement - I : If a, b, c are positive real quantities8 8 8

3 3 3

1 1 1 a b ca b c a b c

Statement -II: Let 1 2 3, , .......... na a a a be n +ve real numbers and let m be real number

then 1 2 1 2.......... .......... mm m nn na a a a a a

n n

if 0,1m R

15 Statement - I : a, b, c are in H.P. and n does not lie between 0 and 1 then an + cn > 2bn.

Statement - II : AM GM HM 16 Statement - I : If ai’ s are all +ve real numbers, then (1 + a1 + a1

2) (1 + a2 + a22) ... (1 + an + an

2) 3n

a1a2 ... an

Statement - II : AM GMMatrix Type Questions17. List - I List - IIA If a,b,c, are in GP , then the equation 2 2 0ax bx c P. AP

and 2 2 0 dx ex f have a common root if , ,d e fa b c are

in


RAMKI

B If a,b,c,d and p an distrinct real numbers such that Q. GP

2 2 2 2 2 2 22 0a b c p ab bc ca p b c d ,

then a,b,c,d are in

C. If 1, 1, 1x y z are in GP then1 1 1, ,

1 n 1 n 1 n l x l y l z are in R.HP

D. Let a,b,c,d are the numbers are in A.P then abc,abd, acd S.AGPbcd are in

Comprehensive Type QuestionsPassage - I

In an increasing G.Pof the first term’a’ and common ratio ‘r’ , the sum of the first and the last termis 66, the prdouct of the second and the last term is 128, and the sum of all the terms is 126.

Answer the following questions18. The value of a is

A) 2 B) 3C) 4 D) 5

19. The common ratio (r) isA) 5 B) 4C) 3 D) 2

20. No of terms of GP isA) 6 B) 7C) 8 D) 9

Passage - II The equation x4 + ax3 + bx2 + cx + 1 = 0 has only real roots , given that a,b and c are non - negative

values, thenAnswer the following questions

21. Minimum value of a isA) 5 B) 4C) 6 D) 7

22. Minimum value of b i sA) 6 B) 7C) 8 D) 9

23. Minimum value of c isA) 3 B) 2C) 4 D) 15

KEY1.(A) 2( B) 3 (B) 4(A) 5(A) 6(D) 7(A) 8(B) 9(B) 10(A)11(A,B,D) 12(A,D) 13 (A) 14(A) 15(A) 16(A)

17

a pb qc Rd R

18(A) 19(B) 20(A) 21(B) 22(A) 23(C)


RAMKI

Hints and Solutions1. Let x be the first term and d be the c. d of A.P.

a = x + (p - 1) d

b = x+ (q-1) d

a bdp q ----------(1)

so, x = a -

p 1 a bp q

pa qa pa pb a b pb qa a b

p q p q

Hence, sp+q=

p q a ba b2 p q

2. Let first term = a c. d = d, Sm = sum of first m terms

Then given Sm = Sm+n – Sm = Sm+p – Sm

m m n m m m nS S S 2S S

m m n2 2a m 1 d 2a m n 1 d2 2

2 2 22a 2m m n d m n 2mn m n 2m 2m

2 22a m n d n m 2mn m n ....... (i)

also 2 Sm = Sm+p

2 22a m p d p m 2mp m p ....... (ii)

from (i) and (ii)

2 2 2 2n m 2mn m n p m 2mp m pm n m p

2mn 2mpn m 1 p m 1m n m p

m p m nm n m p

mp mn

1 1 1 1m n m pm p m n


RAMKI

3. We have, Sk = a + ar + . . . + ark – 1 = ka 1 r

1 r

S1 + S2 + S3 + . . . + Sn = na a

1 r 1 r

(r + r2 + . . . + rn)

=

n n

2

r 1 r ar 1 rna a na1 r 1 r 1 r 1 r 1 r

4. The given sum S = (x + y) + (x2 + xy +y2) +.......

= 1

x y 2 2 3 3 4 4x y x y x y ....

=

2 3 2 31 x x .... y y ....x y

=

2 2 2 2 2 21 x y 1 x y x y xy x y xyx y 1 x 1 y x y 1 x 1 y 1 x 1 y

5. Given 2 a b b cb ac, x , y2 2

Consider a c 2a 2c 2a 2cx y a b b c a ac ac c

a c2 2a c

6. Number of students giving wrong answers to at least r questions = 2p–r

Number of students giving wrong answers to at least (r + 1) questions = 2p–r–1

\ Number of students giving wrong answers to exactly r questions

= 2p–r – 2p–r–1. Also number of students giving wrong answers to exactly p questions

= 2p–p = 20 = 1

\ Total number of wrong answers

p 1 p 2 p z p 3 1 0 01 2 2 2 2 2 ......... p 1 2 2 n 2

p 1 p 2 p 2 p 3 p 3 0 02 2 2.2 2.2 3.2 ........ n 1 2 n.2

= 2p–1 + 2p–2 + 2p–3 + ........ + 20 = 2p – 1

p p 112 1 2047 2 2048 2 p 11


RAMKI

7 Since a, b,c R and distinct

AM GM HM

Since b ac and consider AM and HM of a and c,

a c 2acb2 a c

From first inequality (a + c) > 2b

2a ac 2ab 0

From second inequality b (a + c) > 2ac

2ab 2bc 4ac 0

Adding the two inequalities a2 + 2bc – 3ac > 0

8. Consider the opposite numbers

ax, ax, .......... ky times and by, by, ........ kx times

x x x xa a .........ky times b b .......kx timesAM

kx ky

x y x ykya kxa ya xa

k x y x y

1

x x y y k x yGM a .a ......ky times b .b ........kx times

1 kxy xy

x ky y kx k x y k x y x ya .b ab ab ....... (i)

As 1 1 x y1, 1, i.e., x y xyx y xy

x y x yya xa a ai becomes ab or abxy x y

9 Let us add one more number, an + 1, to the given sequence. The number an + 1 is such that |an + 1| = |an + 1|.Squaring all the numbers, we have

21a 0

2 22 1 1a a 2a 1

2 23 2 2a a 2a 1

2 23 2 2a a 2a 1


RAMKI

2 2n n 1 n 1a a 2a 1

2 2n 1 n na a 2a 1

Adding the above equalities, we get

2 2 2 2 2 2 21 2 n n 1 1 2 n 1 2 na a . . . a a a a . . . a 2 a a . . . a n

2 21 2 n n 1a a . . . a n a n 1 2 na a . . . a 1

n 2

.

10. Keeping in view that a + b + c = 18

a2b3c4 is max. when 2 3 4a b c

2 3 4

is max.

The sum of factors is a b c2. 3. 4.2 3 4 = a + b + c = 18

hence product will be max. when all the factors are equal a b c a b c 18 22 3 4 2 3 4 9

a = 4, b = 6, c = 8

Max. value of a2 b3 c4 is 4263 84.

11. Let 1st term is x and 2 1 thn term = y

1. In .A P 2

nx yt a .

2. in G.P nt xy b

H.P. 2

nxyt c

x y

2b ac and a b c , equality holds good if a b c

12. 1 1 11 .............2 3 2 1

na n

1 1 1 1 11 .......... .........2 2 4 4 4

2 4 2 11 ............2 4 2 1

n

n n


RAMKI

100 100a

again 112 2n

na n

100 100a

13. Let x, y, z be three numbers with weights x, y, z respectively. Then

1

x y z x y zx.x y.y z.z x .y .zx y z

weighted A.M. weighted G.M.

x y z2 2 2x y zx y z x y z

x y z

(i)

Again let 1 1 1, ,x y z be three numbers with wights x, y, z respectively then

1x y z

x y z

1 1 1x.. y. z.1 1 1x y z . .

x y z x y z

weighted A.M. weighted G.M.

x y z

x y z

3 1x y z x y z

x y zx y z x y zx y z

3

(ii)

Using (i) and (ii), we get the result.

14. We know 88 8 8a b c a b c

3 3

(i)

And 13

a b c abc3

62 2 2a b c a b c

3

(ii)

Now (a – b)2 + (b – c)2 + (c – a)2 0

2 2 2a b c ab bc ac


RAMKI

2a b c 3 ab bc ac

2a b c 1 ab bc ac

3 3

From (i) and (ii)

8

2 2 2a b c 1 a b c ab bc ac3 3

from (i) and (iii)

8 8 8

2 2 2a b c 1 a b c ab bc ac3 3

8 8 8

3 3 3

a b c 1 1 1a b ca b c

Equality hold when a = b = c

15.nn na c a c

2 2

, if n does not lie between 0 and 1.

But we know that A.M. > G.M. > H.M.

a c

2

the A.M. of a and c is > b i.e., H.M. of a and c because a, b, c are given to be in H.P..

n

na c b2

nn n

na c a c b2 2

an + cn > 2bn

16. Using A.M. G.M., we get 1 + a1 + a12 3a1 , 1 + a2 + a2

2 3a2, ... 1 + an + an2 3an .

Multiplying, we get (1 + a1 + a12) (1 + a2 + a2

2) ... (1 + an + an2) 3n a1a2 ... an.

17.

a pb qc Rd R

Passage - I

Let a be the first term, r be the common ratio and n be the number of terms

Since, a + a rn–1 = 66 .......... (i)

Also ar. arn-2 = 128


RAMKI

2 n 1a r = 128 .......... (ii)

and

na r 1126

r 1

.......... (iii)

From (i) and (ii), 2

128a 1 66a

2a 66a 128 0 a 64 a 2 0

so, a = 2. Then 4.rn–1 = 128 (from ii)

n 1 nr 32 r 32r

so, from (iii), 32r 1

2 126r 1

32r 1 63r 63 31r 62 r 2

so, n 6r 64 2 n 6

Similarly if a = 64, therefore r = 1/2 and then n = 6( discarded as the given GPs increasing )

Hence number of terms = 6

Passage - IILet f(x) = x4 + ax3 + bx2 + cx + 1As a, b and c are non-negative, no root of the equation f(x) = 0 can be positive. Further asf(0) 0, all the roots of the equation, say x1, x2, x3 and x4, are negative. We have

1 1 2x a, x x b, 1 2 3 1 2 3 4x x x c and x x x x 1 Using A.M. G.M. for positive numbers –x1, –x2, –x3 and –x4, we get

a 1 a 44

Using A.M. G.M. for positive numbers x1 x2, x1 x3, x1 x4, x2 x3, x2x4 and x3x4, we get

b 1 b 66

Finally using A.M. G.M. for positive numbers –x1 x2 x3, – x1 x2 x4, – x1x3x4 and –x2x3x4, we get

c 1 c 44 .


RAMKI

LEVEL-I

Model Questions

Single Correct Answer Type Questions

A.P. G.P. & H.P1. If the sum of first 2n terms of A.P. 2, 5, 8,... is equal to the sum of the first n terms of the A.P. 57, 59, 61,

....., then n equals

(A)10 (B) 12

(C) 11 (D) 13

2. Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second

term is 34

, then

(A) 7 3a ,r4 7

(B) 3a 2,r8

(C) 3 1a ,r2 2

(D) 1a 3,r4

3. If S denotes the sum to infinity and Sn the sum of n terms of the series 1 1 11 .....,2 4 8

such that

S – Sn 1

1000 , then the least value of n is

(A) 8 (B) 9

(C) 10 (D) 11

4. If 2 3x 1 a a a ...to (| a | 1) and 2 3y 1 b b b ...to (| b | 1) , then

2 2 3 31 ab a b a b .... is equal to

(A) xy

x y 1 (B) x y

x y 1

(C) x y

x y 1

(D) none

5. If a1, a2, a3, ..... is an A.P. such that

1 5 10 15 20 24a a a a a a 225 ,

then 1 2 3 23 24a a a ... a a is equal to

(A) 909 (B) 75 (C) 750 (D) 900


RAMKI

6. If the sum of first p terms, first q terms and first r terms of an A.P. be a, b and c respectively, then

a b cq r r p p qp q r

is equal to

(A) 0 (B) 2

(C) pqr (D) 8abcpqr

7. The numbers 32sin2q–1, 14, 34–2 sin2q form first three terms of an A.P. Its fifth term is equal to

(A) –25 (B) –12

(C) 40 (D) 53

8. If sum of n terms of an A.P. is 3n2 + 5n and Tm = 164, m = ?

(A) 26 (B) 27

(C) 28 (D) none of these

9. If the sum of the series 2 35 25 1252 ......... isx x x

finite, then

(A) |x| > 5 (B) – 5 < x < 5

(C) |x| < 5/2 (D) |x| > 5/2

AM GM & HM10. Between 1 and 31 are inserted m arithmetic means so that the ratio of the 7th and (m–1)th means is 5 :

9. Then the value of m is

(A) 12 (B) 13

(C) 14 (D) 15

SUMMATION OF SERIES

11. If 1 1 1 1 11 .......3 5 7 9 11 4

, then value of

1 1 1 .......... is1.3 5.7 9.11

(A) 8

(B) 6

(C) 4

(D) 36

INEQUALITY12. If three positive real numbers a, b, c are in A.P., with abc = 4, then the minimum value of b is

(A) 41/3 (B) 3

(C) 2 (D) 1/2


RAMKI

13. If x, y and z are positive real numbers such that x + y + z = a then

(A) 1 1 1 9x y z a (B)

1 1 1 9x y z a

(C) 38a x a y a z a27

(D) none of these

14. If a, b and c are distinct positive real numbers and a2 + b2 + c2 = 1, then

ab + bc + ca is

(A) less than 1 (B) equal to 1

(C) greater than 1 (D) any real number

15. If a, b and c are positive real numbers, then least value of

(a + b + c) 1 1 1a b c

is

(A) 9 (B) 3

(C) 10/3 (D) none of these

More Than One Correct Answer Type QuestionsA.P G.P & H.P

16. The sum of the first n term (n > 1) of an A.P. is 155 and the common difference is 2. If the first term is apositive integer, then

(A) n can not be even (B) n can not be odd

(C) 5 (D) 6

17. If b1, b2, b3 (b1 > 0) are three successive terms of a G.P. with common ratio r, the value of r for which theinequality b3 > 4b2 – 3b1 holds is given by

(A) r > 3 (B) r < 1

(C) r = 3.5 (D) r = 5.2

18. Let S1, S2, ............. be squares such that for each n 1 , the length of a side of Sn equals to the lengthof a diagonal of Sn+1. If the length of a side of S1 is 10 cm, then for which of the following value (s) of n isthe area of Sn less than 1 sq. cm?

(A) 7 (B) 8

(C) 9 (D) 10

19. All the term of an A. P. are natural numbers and the sum of the first 20 terms is greater than 1072 andless than 1162. If the sixth term is 32 then

(A) first term is 12 (B) first term is 7

(C) common difference is 4 (D) common difference is 5

20. Let the sets A = {2, 4, 6, 8, .....} and B = {3, 6, 9, 12, .....}, and n(A) = 200, n(B) = 250. Then

(A) n(A B) 67 (B) n(A B) 450

(C) n(A B) 66 (D) n(A B) 384


RAMKI

21. In a GP the product of the first four terms is 4 and the second term is the reciprocal of the fourth term. Thesum of the GP up to infinite terms is

(A) 8 (B) -8

(C) 8/3 (D) –8/3

22. Let x1, x2, ............ be positive integers in A.P., such that x1 + x2 + x3 = 12 and x4 + x6 = 14. Thenx5 is

(A) a prime number (B) 11

(C) 13 (D) 7

SUMMATION OF SERIES23. The sum of the numerical series

1 1 1 .........upto n terms3 7 7 11 11 15

, is

(A) 3 4n 3

4

(B) 4

3 4n 3

(C) less than n (D) greater than n / 2

24. The sum of the series 1 3 7 15 .............2 4 8 16

(A) upto n terms is : n

1n 12

(B) upto n terms is : n

1n2

(C) upto n terms is 2 (D) upto infinity is infinity

25. If n

r 1r r 1 2r 3

= an4 + bn3 + cn2 + dn + e, then

(A) a + c = b + d (B) e = 0

(C) a, b – 2/3, c –1 are in A.P. (D) c/a is an integers

(Assertion – Reason Type Questions)Each question contains STATEMENT – 1 (Assertion) and STATEMENT – 2 (Reason). Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Instructions:(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation forStatement-1(B) Statement-1 is True, Statement-2 is True; Statement–2 NOT a correct explanation forStatement-1.(C) Statement–1 is True, Statement–2 is False(D) Statement –1 is False, statement–2 is True.


RAMKI

26. For 2 r 1rr 1, and x 1 Let t 1 2x 3x ......... rx

STATEMENT-1 : Sum of t1 + t2 + ....... + tn is

n 1 n

2 3

n 1 x 2x 1 x

1 x 1 x

STATEMENT-2 : For r

2 r 1 1 xr 1, and x 1,1 x x .......... x1 x

and

r r2 r 1

2

1 x rx1 2x 3x .......... rx1 x1 x

27. STATEMENT-1 : If x > 1, the sum to infinite number of

2 31 1 11 3 1 5 1 7 1 ........

x x x

is 2x2 – x

STATEMENT-2 : If 0 < y < 1, the sum of the series 2 3

2

1 y1 3y 5y 7y ........is1 y

28. Let a, r Î R – {0, 1, – 1} and n be an even number.STATEMENT-1 : a. ar. ar2 . . . arn – 1 = (a2 rn – 1)n/2.STATEMENT-2 : Product of kth term from the beginning and from the end in a G.P. is independent of k.

Comprehension Type QuestionsThis section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered.Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

P29–31 : Paragraph for Question Nos. 29 to 31An A.P. is a sequence whose terms increase or decrease by a fixed number, called the common differenceof the A.P.. If ‘a is the first term and ‘d’ the common difference, the A.P. can be written as a, a+d, a +2d,…… The nth term an is given by an = a + (n – 1)d. The sum sn of the first n terms of such an A.P. is

given by : sn = n2 (2a + (n – 1)d) =

n2 (a + l) where l is the last term (i.e., the nth term of the A.P. ). If a,

b, c are in A.P., then a cb

2

is the A.M. of a and c. The n numbers AA1, A2,….., An are said to be A.M.’s

between the numbers a & b if

a, A1, A2, …..,An, b are in A.P. If ‘d’ is the common difference of this A.P., then b adn 1

r(b a)A a rn 1

, where AAr is the rth mean

29. If 6 A.M.’s are inserted between 1 and 9/2, then the 4th arithmetic mean is equal to

(A)32 (B)3

(C) 2/3 (D)13/5


RAMKI

30. If log 2, log (2x – 1) and log (2x + 3) are in A.P., then the value of x is(A)5/2 (B)log25(C)log35 (D)log53

31. If am be the mth term of an A.P., then 2 2 2 2 2 21 2 3 4 2n 1 2na a a a ...... a a is equal to

(A) 2 21 2n

2n 1 a an

(B) 2 21 2n

n a a2n 1

(C) 2 21 2n

n a a2n 1

(D) 2 21 2n

2n 1 a an

P32–34 : Paragraph for Question Nos. 32 to 34If x1, x2,.. . . . . xn are ‘n’ positive real numbers; then A.M. ³ G.M. ³ H.M.

1/ n1 2 n1 2 n

1 2 n

x x .......x n(x x .......x ) 1 1 1n ......x x x

equality occurs when numbers are same using

this concept.32. If a > 0, b > 0, c > 0 and the minimum value of a(b2 + c2) + b(c2 + a2)+ c(a2 + b2) is labc, then l is

(A) 1 (B) 2(C) 3 (D) 6

33. If a, b, c, d, e, f are positive real numbers such that a + b + c + d + e + f = 3, thenx = (a + f)(b + e)(c + d) satisfies the relation(A) 0 < x £ 1 (B) 1 £ x £ 2(C) 2 £ x £ 3 (D) 3 £ x £ 4

34. If a and b are two positive real numbers, and a + b = 1, then the greatest value of a3b4 is

(A) 2 33 475

(B) 3 4

7

3 47

(C) 7

3 4

73 4

(D) none of these

Matrix–Match Type QuestionsThis section contains 2questions. Each question contains statements

given in two column which have to be matched. Statements (A, B, C, D)

in Column I have to be matched with statements (p, q, r, s) in Column II.The answers to these questions have to be appropriately bubbles

as illustrated in the following example.

If the correct matches are A–p, A–s, B–q, B–r, C–p, C–q and D–s,

then the correctly bubbled 4 × 4 matrix should be as follows :35. If a, b, c are in H.P. then

Column I Column II

(A) a b c, ,

b c a c a b a b c (p) H.P.

(B) 1 1 1, ,

b a b b c (q) G.P.


RAMKI

(C) b b ba , ,c2 2 2

(r) A.P.

(D) a b c, ,

b c c a a b (s) A.G.P.

36. Let a, b, g be distinct real numbers which are in A.P. If a, b, g are the roots of

x3 + bx + c = 0, then

Column I Column II

(A) b (p) 0

(B) c (q) negative

(C) 2b c (r) positive

(D) b2 + c2 (s) non-negative

(t) nonpositive

Practice Questions

SINGLE ANSWER TYPE QUESTIONS

37. The third term of a G.P. is 4. The product of the first five terms is

(A) 43 (B) 45


38. If a, b, c, d are in H.P,, then ab + bc + cd is equal to

(A) 3 ad (B) (a + b) (c + d)

(C)3 ac (D) none of these

39. If S be the sum, P the product and R the sum of the reciprocals of n terms of a G.P., then

nSR

(A) P (B) P2

(C) P3 (D) P

40. If x, y, z be respectively the pth, qth and rth terms of G.P., then

q r log x + (r – p) log y + (p – q) log z =

(A) 0 (B) 1

(C) –1 (D) none

41. The sum of integers from 1 to 100 which are divisible by 2 or 5 is

(A) 300 (B) 3050

(C) 3200 (D) 3250


RAMKI

42. The interior angles of a polygon are in arithmetic progression. The smallest angle is 1200 and the commondifference is 5. The number of sides of the polygon is

(A) 7 (B) 9

(C) 11 (D) 16

43. If Sn= nP + 12

n (n – 1) Q where Sn denotes the sum of the first n terms of an A.P., then the common

difference is

(A) P + Q (B) 2P + 3Q

(C) 2Q (D) Q

44. If a1, a2, a3 ….. an are in H.P., then 1 2 n

2 3 n 1 3 n 1 2 n 1

a a a, ,...a a ...a a a ... a a a ... a are in

(A) A.P. (B) G.P.

(C) H.P. (D) none of these

45. If A and G be the A.M. and G.M. respectively between two numbers, then the numbers are

(A) 2 2A G A (B) 2 2A A G

(C) 2 2A A G (D) 2 2G A G

46. If x > 0 and 8 1642 2 2 2 2log x log x log x log x log x ........ 4 then x equals

(A) 2 (B) 3

(C) 4 (D) 5

47. If one geometric mean G and two arithemtic means p and q be inserted between two numbers, then G2 isequal to:

(A) (3p – q) (3q – p) (B) (2p – q) (2q – p)

(C) (4p – q) (4q – p) (D) none of these

48. If the arthimetic mean between a and b is n 1 n 1

n n

a b ,a b

then n =

(A) 0 (B) 1

(C) –1 (D) 1/2

49. The sum upto (2n + 1) terms of the series

a2 – (a + d)2 +(a + 2d)2 – (a + 3d)2 + ........ is

(A) a2 + 3nd2 (B) a2 + 2nad + n (n – 1)d2

(C) a2 + 3nad + n (n – 1)d2 (D) none of these

50. The positive integer n for which 2 × 22 + 3 × 23 + 4 ×24 + .............. + n × 2n = 2n+10 is

(A) 510 (B) 511

(C) 512 (D) 513


RAMKI

51. If a, b and c are positive real numbers then a b cb c a is greater than or equal to

(A) 3 (B) 6

(C) 27 (D) none of theseMore Than One Correct Answer Type Questions

52. The numbers sin x

6 , cos x and tan x will be in G.P. if

(A) x3

(B) 5x6

(C) x 2K3

(D) 2K6

53. If the system of linear equation x + 2ay + az = 0, x 3by bz 0, x + 4 cy + cz = 0, has a non-zerosolution, then a,b, c

(A) can all be zero (B) are in A.P.

(C) are in G.P. (D) are in H.P. if abc 0

54. The next term of the geometric progression x, x2 + 2, x3 + 10 is

(A) 0 (B) 54

(C) 72916

(D) 16729

55. If sum of n terms of an A.P. is given by Sn = a + bn + cn2 where a, b, c are independent of n, then

(A) a = 0 (B) common difference of A.P. must be 2b

(C) common difference of A.P. must be 2c (D) All above

56. If a, b, c are in A.P., then 2ax+1, 2bx+1, 2cx+1, x 0 are in

(A) A.P. (B) G.P. when x > 0

(C) G.P. if x < 0 (D) G.P. for all x 0

(Assertion – Reason Type Questions)Each question contains STATEMENT – 1 (Assertion) and STATEMENT – 2 (Reason). Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Instructions:(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation forStatement-1(B) Statement-1 is True, Statement-2 is True; Statement–2 NOT a correct explanation forStatement-1.(C) Statement–1 is True, Statement–2 is False(D) Statement –1 is False, statement–2 is True.

57. STATEMENT-1 : If the sides of a right angled triangle are in G.P. then the common ratio of the G.P. cantake two and only two values

because

STATEMENT-2 : The common ratio is either greater than 1 or less than 1


RAMKI

58. STATEMENT-1 : If all terms of a series with positive terms are smaller than 10–5, then the sum of theseries upto infinity will be finite

STATEMENT-2 : If n 5nS

10 then lim

n nS is finite

59. Statement–1 : In the expression (x + 1) (x + 2) . . . (x + 50), coefficient of x49 is equal to 1275.

Statement–2 : n

r 1

n n 1r , n N

2

.

Comprehension Type QuestionsThis section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered.Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.P60–62 : Paragraph for Question Nos. 60 to 62

The sum of the squres of three distinct real numbers, which are in G.P., is S2. If their sum is a S thenanswers the following questions.

60. a2 lies

(A) (1/3, 2) (B) (1, 2)

(C) (1/3, 3) (D) none of these

61. If a2 = 2, then value of r equals

(A) 1 5 32

(B) 1 3 52

(C) 1 5 32

(D) 1 3 53

62. If we drop the condition that the G.P. is strictly increasing and take a2 = 3, then common ratio is givenby

(A) 2 (B) +1

(C) 0 (D) 3

Matrix–Match Type Questions

This section contains 2questions. Each question contains statements


in Column I have to be matched with statements (p, q, r, s) in Column II.

The answers to these questions have to be appropriately bubbles



then the correctly bubbled 4 × 4 matrix should be as follows :


RAMKI

63. Sum of the series upto n terms

Column I Column II

(A) 13 + (1.5)3 + 23 + (2.5)3 + ...... (p) 1 n n 1 n 2 3n 512

(B) 1(2)2 + 2(3)2 + 3(42) + .......... (q) n (3n2 + 6n + 1)

(C) (n2 – 1)2 + 2 (n2 – 22) + (r) 2 21 1n 1 n 232 8

3(n2 – 32) + ......

(D) (2) (5) + (5) (8) + (8) (11) + ...... (s) 221 n n 14

64. Match the value of x on the left with the value on the right.

Column I Column II

(A) 52 54 56 .......52x = (0.04)–28 (p) 3 log3 5

(B) 51 1 1log .............2 4 8 16x 0.2

(q) 4

(C) 25 2 31 1 1log ..........3 3 3x 0.16

(r) 2

(D) 3x–1 + 3x–2 + 3x–3 + .............. (s) 7

22

1 12 5 5 1 ..........5 5

(t) even integer

LEVEL-I

(ANSWER KEY)

1-C 2-D 3-D 4-A 5-D 6-A

7-D 8-B 9-A 10-C 11-A 12-A

13-A 14-A 15-A 16-A,C 17-A,B,C,D 18-B,C,D

19-B,D 20-C,D 21-A,B,C,D 22-A,D 23-A,B,C,D 24-A,D

25-A,B,C,D 26-A 27-A 28-B 29-B 30-B

31-B 32-D 33-A 34-B

35- A P BR CQ DP


RAMKI

36. A Q BP C R DR

37-B 38-A 39-B 40-A 41-B 42-B

43-D 44-C 45-B 46-C 47-B 48-A

49-D 50-D 51-A 52-A,C 53-A,D 54-B,C

55-A,C 56-B,D 57-B 58-B 59-A 60-C

61-B 62-B

63. A R BP CS DQ

64. A S BR,T C Q,T DP

29-91 30-2

HINTS AND SOLUTIONS

Level-I

(Model Questions)

1. S2n = Sn’

2n n2.2 2n 3 2.57 n 1 22 2

5n 55

2. Given a 4

1 r

and

3ar4

Eliminating a, we have 16r2 – 16r + 3 = 0

or (4r – 3) (4r – 1) =0

r = 3/4, 1/4.

Hence r = 1/4 so that a = 3.

r = 3/4 is not given in any of the four choices so we choose r = 1/4 only.

3. 1S 2

1 1/ 2

n

n n 1

1 1/ 2 1S1 1/ 2 2

=2- n 1

12


RAMKI

n n 1

1 1S S10002 or n 12 1000

Now 210 = 32 × 32 = 1024

\ n – 1 10 or n 111

Hence the least value is 11.

4. 1 1x .y

1 a 1 b

x 1 y 1a .bx y

2 21 ab a b ...

1 1 xy

x 1 y 11 ab x y 11

xy

5. We given, 5 20 1 24 10 15 1 24a a a a ,a a a a

Hence the given relations reduce to, 1, 243 a a 225 , giving 1 24a a 75

Hence S24 = 1 24n a 24 / 2 a a 12 75 9002

l

6. ppS [2A p 1 d] a2

2a 2A p 1 dp

.... (i)

2b 2A q 1 dq

.... (ii)

2c 2A r 1 dr

.... (iii)

Multiply (i), (ii) and (iii) by q – r, r – p and p – q respectively and add

a q r 0p

7. Since the numbers are in A.P.

\ 28 = 32 sin 2q –1 + 34 - 2 sin 2q


RAMKI

sin2

sin2

9 81or 283 9

, where x = 9sin 2 q

or 2x 84x 243 0

or (x – 81) (x – 3) = 0 x 81 or 3

sin2 2 1/2x 9 81,3 or 9 ,9

sin2 2 or1/ 2

since sin 2q cannot be greater than 1 so we choose sin 2q = 12

Hence the terms in A.P. are

03 , 14, 27 i.e. 1, 14, 27.

5T a 4d = 1 + 4.13 = 53

8. Tm = Sm – Sm – 1

\ 164 = 3 (2m – 1) + 5.1 \ 6m = 162

9. We can rewrite the series as

2 35 5 51 1 ..........x x x

We can sum up this series if |5/x| < 1

| x | 5

10. 1 2 m1,x ,x ...x , 31 is an A.P. of (m + 2) terms.

31 = Tm+2 = a + (m + 1) d

= 1 + (m + 1)d

30d

m 1

Now 7

m 1

x 5x 9

8

m

T a 7d 5T a (m 1)d 9

Now put for a and d and we get m = 14.

11. We have

1 1 1 1 11 .........4 3 5 7 9 11


RAMKI

2 2 2 .........1.3 5.7 9.11

1 1 1 ..........1.3 5.7 9.11 8

12. Using AM GM

13

a b c abc3

13

3b abc3 (since 2b = a + c)

13b 4

13. Since AM HM

x y z 31 1 13x y z

a 3 1 1 1 9or1 1 13 x y z ax y z

14. Since a and b are unequal, 2 2

2 2a b a b2

(A.M. > G.M. for unequal numbers)

2 2a b 2ab

Similarly b2 + c2 > 2bc and c2 + a2 > 2ca

Hence 2 (a2 + b2 + c2) > 2 (ab + bc + ca)

ab bc ca 1

15. Using A.M. G.M.

1/3

1/31 1 1 1 1 1a b c abc and3 3 a b c abc

1 1 1 1 1a b c . 13 3 a b c


RAMKI

1 1 1a b c 9a b c

Equality will hold when a = b = c

16. Let the numbers in A.P. are a, (a + 2), (a + 4)....... a + (a + 2) + (a + 4) + ........ = 155

n 2a n 1 2 1552

n a n 1 155

n can not be even as sum is an odd number

17. We have

b3 > 4b2 – 3b1 2

1 1 1b r 4b r 3b

2r 4r 3 1b 0

2r 4r 3 0 r 3 r 1 0

r 3 or r 1

Since r = 3.5 and r = 5.2 are both greater than 3, (c) and (d) are also true.

18. Let an denotes the side of the square Sn then

n n 1a 2a

n 1

n

a 1a 2

n 1

n 11a a2

(G.P. formula) = 10n 1

12

Now, we must have 2na 1

2n 21100 12

n2 200 n 8

19. We have 1072 < 10 (2a + 19d) < 1162 and a + 5d = 32

1072 < 640 + 90d < 1162

\ 432 522d90 90

and d is natural number, so d = 5 a = 7


RAMKI

20. We have to find common terms in 2, 4, 6, 8,.... to 200 terms & 3, 6, 9, 12, ...... to 250 terms

last term of the first AP = 2 + (200 – 1) 2 = 400

last term of second AP = 3 + (250 – 1) 3 = 750

\ We will get common term upto 400.

\ In first AP common difference = 2

\ In second AP common difference = 3

\ common difference of the common AP = 2 × 3 = 6

\ The common terms are 6, 12, 18, 24........ 396.

\ Let number of terms = n

\ 396 = 6 + (n – 1) 6

\ 396 = 6n n 66

n A B = 66

n A B = n A n B n A B

= 200 + 250 – 66 = 384

21. Let the first four terms be 33

a a, ,a,arrr

then

33

a a a ar 4rr

4a 4

a 2

Also given 3

a 1r ar 2 1r

2

1r2

so, the sum to infinite series 3

2

arS

1 r

( first term = 3

ar

and common ratio = r2)

22. Let the positive integers in A.P. are a, a + d, a + 2d,....

according to question a + a + d + a + 2d = 12

a + d = 4 .... (i)

also a + 3d + a + 5d = 14

a + 4d = 7 .... (ii)

From (i) and (ii) a = 3, d = 1

Hence x5 = a + 4d = 7


RAMKI

23. (a), (b) on rationalizing each term, we get series

7 3 11 7 15 11 .........upto n term7 3 11 7 15 11

1 3 4n 34 which is equal to

n3 4n 3

choices (a) and (b) are correct

c) Since n n

3 4n 3

, choice c is also correct

d) And n n n

23 4n 3 4n

24. n 2 3 n

1 1 1 1S 1 1 1 ......... 12 2 2 2

2 n n

1 1 1 1n ......... n 12 2 2 2

As n

1n , 0, Hence S2

25. an4 + bn3 + cn2 + dn + e

n n

r 1 r 12 r r 1 r 2 r r 1

2 1n n 1 n 2 n 3 n n 1 n 24 3

4 3 21 3n 16n 27n 14n6

26. Ans: (A)

According to theory

r r

r 2

1 x rxt1 x1 x

n n nr r

r 2r 1 r 1 r 1

1 1t 1 x rx1 x1 x


RAMKI

n n n 1

2 2

x 1 x x 1 x1 1 nxn1 x 1 x 1 x1 x 1 x

n 1 n

2 3

n 1 x 2x 1 x

1 x 1 x

27. Ans: (A)

Let

S = 1 + 3y + 5y2 + 7y3 + ........... (i)

yS = y + 3y2 + 5y3 + .......... (ii)

Subtracting (2) from (1) we get

2 3 2y1 y S 1 2y 2y 2y ......... 11 y

2 2

1 2y 1 yS1 y 1 y 1 y

Putting 1y 1 , we getx

2

2

1 1 1/ xS 2x x1/ x

.......... (iii)

28. Ans:(B)

Statement (1) a. ar. ar2 ..............arn-1 = an ( ) /

n n 1 n 22 n 12r a r

statement (1) is correctStatement (2)from beginning Tk = ark-1

from the end TK = Tn–K+1 from beginning = arn-k

Product of Tk from beginning and Tk from end = ark-1 ́ arn-k = a2rn-1

Which is independent of ‘k’So, statement (2) is correct but not explanation for statment (1)

29. Let thr A.M. between a and b is Ar

r(b a)A a rn 1

AA4= 1+4

9 126 1

=1 + 2 = 3

30. As log 2, log (2x – 1) and log (2x + 3) are in A.P. thenlog (2x-1) -log2 = log (2x+3) - log (2x-1)

or, x x

x2 1 2 3log log

2 2 1


RAMKI

or , x x

x2 1 2 3

2 2 1

or 2x x2 4.2 5 0

or, x x2 5 2 1 0

as, 2x +1 0

x2 5 0 or, x2 5

or, x = log2 5

31. Let c.d of the A.P. a1, a2, a3, ---------a2n-1, a2n is ‘d’a2 - a1 = a4 – a3 = ---------- = a2n-a2n-1 =d

Now 2 2 2 2 2 21 2 3 4 2n 1 2na a a a ...... a a = - d 1 2 3 4(a a a a + ............+a2n-1+a2n)

1 2n 1 2n2nd a a d a a n2

or, 2 2 2 2 2 21 2 3 4 2n 1 2na a a a ...... a a =

2n 1 1 2na a a a n2n 1

= 2 2

1 2na a n

2n 1

32. As A.M. G.M.

16

b c c a a bc b c a a bc b a c b a

6 b c a c b a

or, 2 2 2 2 2 2b c c a a b 6bc ca ab

or, 2 2 2 2 2 2a b c b c a c a b 6abc

minimum value of 2 2 2 2 2 2a b c b c a c a b is 6 abc

according to question labc = 6abc l= 6

33. As a, b, c,d, e, f, are (+) ve(a+f) (b +e) (c+d) > 0So, X > 0 ...........(1)As A. M. G.. M.

13

a f b e c da f b e c d

3

or, 3 3x3

or, x 1................(2)from (1) and (2) 0 < x 1


RAMKI

34. By weightage mean

3 4a b3 4

a b3 4 77 3 4

or, 3 4

3 4a b a b7

7 3 4

or, 3 4

3 4a b 17

73 4

or,3 4

3 4 7a b 13 4 7

or, a3b4 3 4

73 47

greatest value of a3 b4 is 3 4

73 47

35. (A) a, b, c, are in H.P. , ,1 1 1a b c are in A.P..

multiply each term by a +b+c then substract 2 from each term we get

, ,b c a a c b a b ca b c

are in A.P..

or, , ,a b cb c a a c b a b c

are in H.P..

(B) as, a, b, c are in H.P.

2acba c

.....(1)

Now, b - a =

( )ac 2a a c aa c a c

( )1 a c

b a a c a

similarly ( )1 a c

b c c a c

1 1 a c 2b a b c ac b

(From (1))

, ,1 1 1b a b b c are in A.P..

(C) as a, b, c are in H.P.

2acba c


RAMKI

b ac2 a c

-----(i)

2b aa2 a c

and 2b cc

2 a c

2 2b b ac ba c2 2 a c 2

Hence , ,b b ba c2 2 2

are in G.P..

(D) as a, b, c are in H.P.

, ,1 1 1a b c

are in A.P..

Multiply each term by a + b +c than substract 1 from each term we get

, ,b c a c a ba b c

are in A.P..

or, , ,a b cb c c a a b

are in H.P..

36. Ans:A-q, B-p, C-r, D-r

0 3 0 0

\ c = 0

Also, (a + g) b + ag = b

2b 0

Thus, 2 2 2 2 2b c b 0 and b c b 0

Practice Questions

37. 23T ar 4 ----------(i)

1 2 3 4 5T .T .T .T .T = a.ar.ar2.ar3.ar4 55 10 2 5a r ar 4 , from (i)

38.2acba c

and 2bdcb d

2ac 2bda c b d . 4adb c

ab bc cd 3ad

39. S = a + ar + ar2 + ar2 + ... + arn-1

i.e. n terms


RAMKI

na 1 rS

1 r

..... (i)

\ P = product = a. ar. ar2....arn–1

n n 1 / 2n 1 2 3 4 5 ....n 1 na r a .r

n n 12 2np a r ..... (ii)

2 3 n 1

1 1 1 1 1R ....a ar ar ar ar (n terms)

nn

n 1

11 r 11 1rR .a 1 1/ r r 1 ar

..... (iii)

n

n 1n 1 2n

1 r r 1S a. . .ar a rR 1 r r 1

by (i) and (ii).

n n n 12n 2S /R a r p by (ii)

40. p 1pT AR x

logx logA (p 1)logR

Similary write log y , log z

Multiply by q – r, r – p and p – q and add we get,

q r log x r p log y p q logz 0

41. L.C.M. of 2 and 5 is 10.

Numbers divisible by 2 will contain numbers which are also divisible by 10. Similarly numbers divisible by5 will contain numbers which are also divisible by 10. Thus the number divisible by 10 will occur twice.Hence we can

write S = S2 + S5 – S10

Now, S2 = 2.50.51 2550

2 by

n n 1n

2

Similarly, S5 = 1050, S10 = 550

S = S2 + S5 – S10

= 2550 + 1050 – 550 = 3050


RAMKI

42. 0nS 2n 4 n 2 180

2

(formula for polygon)

a = 1200, d = 50

nnS [2a n 1 d]2

for A.P..

2n 25n 144 0

n = 9, 16

But when n = 16 then T16 = 1950 which is not possible

n = 9 only

43. . n n n 1t S S P n 1 Q

or nt nQ P Q (linear) \ A.P..

d = coeff. of n = Q and a = P for n = 1

44. 1 2 3 na ,a ,a ........a are in H.P..

1 2 3 n

1 1 1 1, , ......a a a a are in A.P..

Multiply each term by 1 2 3a a a ........ + na then substract 1 from each term

we get2 3 n 1 3 n 1 2 n 1

1 2 n

a a .... a a a .... a a a .... a, ,.....a a a

are in A.P..

1 2

2 3 n 1 3 n

a a,a a .... a a a ... a

,.....

n

1 2 n 1

aa a .... a in H.P..

45. a, A, b in A.P.

2A a b .......... (i)

and a, G, b in G.P.

2G ab .......... (ii)

Solving equation (i) and (ii)

we get a & b equal to A 2 2A G

46. We can write the given equation as1 1 1 11 ...........2 4 8 16

2log x 4

2 2 42log x 4 x 2 x 4


RAMKI

47. Let numbers be a & b

a,a,b in G.P. and a, p, q, b in A.P..

2a ab & p – a = q – p = b – q

we get a = 2p – q & b = 2q – p

so a2 = (2p – q) (2q – p)

48. on putting n = 0, we get a b

2

49. We can write the sum upto (2n + 1) terms as

[a + (a + d)] (–d) + [(a + 2d) + (a + 3d)] (–d) + ....... [(a +(2n – 2) d) + (a + (2n – 1 d]

(–d) + (a + 2nd)2

= (–d) [a + (a + d) + (a + 2d) + ........ + a + (2n – 1) d] + (a + 2nd)2

22nd a a 2n 1 d a 2nd2

= –2nad – n (2n – 1)d2 + a2 + 4n (ad) + 4n2d2

= a2 + 2nad + n (2n + 1) d2

50. We have

2n+10 = 2.22 + 3.23 + 4.24 + ....... + n.2n

n 10 3 4 n n 12 2 2.2 3.2 ........ n 1 .2 n.2

Subtracting, we get

n 10 2 3 4 n n 12 2.2 2 2 ...... 2 n2

n 2n 1

8 2 18 n.2

2 1

n 1 n 1 n 1 n 18 2 8 n.2 2 n 2

102 2n 2 n 513

51. Using A.M. G.M

1/31 a b c a b c a b c 33 b c a b c a b c a

52.sin x .cos x, tan x

6 are in G.P..

2 sin x.tan xcos x6


RAMKI

3 26cos x cos x 1 0

Put cos x = t 3 26t t 1 0

22t 1 3t 2t 1 0

As the quadratic factor has imaginary roots.

\ t = 1/2 i.e.., cos x = 1/2

x 2k3

53. For non-zero solution

1 2a a2ac1 3a b 0 ba c

1 4c c

Hence a, b, c are in H.P. if abc 0.

If abc = 0, then a = b = c = 0

54.2 3

2

x 2 x 10 1x 2,x 2x 2

Given G.P. becomes

2, 6, 18, 45,... or 1 9 81 729, , , .....2 4 8 16

Next term is 72916

or 54.

55. 2n

nS 2a n 1 d a bn cn2

2n n 1 dna a bn cn

2

22d n da n a bn cn

2 2

On comparing coefficient, we get

d da 0,b a ,c2 2

56. Take a = 1, b = 2. c = 3 (As a, b, c are in A.P.)

We get, bx 12 is G.M. of ax 12 and cx 12 for x 0.


RAMKI

57. Ans: (B)

Sol. Let a, ar, ar2 be the sides of right triangle\ a2 + (ar)2 = (ar2)2

4 2r r 1 0 Let r2 = xx2 – x – 1 = 0

1 5x2

2 21 5 1 5r r2 2

r can take only two values A is trueAlso r 1 , because a, a, a can not be the sides of right triangle , therefore either r isgreater than one or less than one.

58.Ans :(B)

Sol Assertion A is false , since each term of the series ..........6 6 6 61 1 1 1

10 10 10 10 is

smaller than 10–5, but its sum upto infinity is infinity

Reason R is false, since 5n

10 is not finite as n

59. Statement -1 Co-efficient of x49 is , equal to

1 + 2 + 3 +........+ 50 =

50 51 25 51

2 =1275

Statement (2) is correct and correct explanation for statement (1)

60. . Let terms of G.P. are a ,a,arr

Given a a ar sr

or 21 r ra

r

= as ..... (i)

and 2

2 2 2 22

a a a r sr

or 4 2

2 22

r r 1a sr

...... (ii)


RAMKI

on squaring (i) & dividing by (ii)

we get .2

22

r r 1r r 1

....... (iii)

. by (iii), 2 2 2 2r 1 r 1 1 0

Since increasing G.P. D 0

21 33

61. If 2 2 then 2 2r r 1 2r 2r 2

3 5r2

62. If 2 3

2 2r r 1 3r 3r 3

22r 4r 2 0 r 1

63. A-r, B-p, C-s, D-q

Sol: A) 3r

1t r 18

n n n 1

3 3r

r 1 r 1 r 1

1 1 1t r 1 r8 8 8

2 21 1n 1 n 232 8

B) tr = r (r + 1)2 = r3 + 2r2 + r

n

22r

r 1

1 1 1t n n 1 n n 1 2n 1 n n 14 3 2

21 n n 1 3n 11n 1012

1 n n 1 n 2 3n 512

C) tr = r (n2 – r2)

n n n2 3

rr 1 r 1 r 1

t n r r


RAMKI

2

22n 1n 1 n n n 12 4

2 21 n n 14

D) tr = (3r – 1) (3r + 2) = 9r2 + 3r – 2

n

rr 1

9 3t n n 1 2n 1 n n 1 2n6 2

3 n n 1 2n 1 1 2n2

= 3n (n2 + 2n + 1) – 2n

= (3n2 + 6n + 1)n

64. A-s, B-r, C-q, D-p

Sol: A) 52+4+6 + ........ 2x = (25)28

x x 1 565 5

2x x 56 0 x 7as x 0

B) 5 55

1/ 42log x log log 0.21 1/ 2

55

1 1log log2 5

5

55

1log2 log 4

log 5

x 2

C) 2.51/ 3log x log log 0.16

1 1/ 3

= log5/2 (1/2) log (2/5)2

= log 4

x 4

D) 2

x2 51/3

31 1/3 1 1/5


RAMKI

x 31 13 52 2

3x 3 log 5

LEVEL-II

Model Questions


1. In a G.P., T2 + T5 = 216 and T4 : T6 = 1:4 and all terms are integers, then its first term is

(A) 16 (B) 14

(C) 12 (D) none

2. If 1 2 3 na ,a ,a ,..a be an A.P. of non-zero terms then

1 2 2 3 n 1 n

1 1 1...a a a a a a

=

(A) 1 n

aa a (B)

1 n

na a

(C) 1 n

n 1a a

(D) none

3. If the ratio of sum of m terms and n terms of an A.P. be m2: n2, then the ratio of its mth and nth terms willbe

(A) 2m – 1 : 2n – 1 (B) M : n

(C) 2m + 1 : 2n + 1 (D) none

4. The ratio between the sum of n terms of two A.P.’s is 3n + 8 : 7 n + 15. Then the ratio between their 12th

terms is

(A) 5 : 7 (B) 7 : 16

(C) 12 : 11 (D) none

5. If A.M. between two numbers is 5 and their G.M. is 4, then the H.M. will be:

a) 165 b)

145 c)

115

d) none of these

6. The A.M. between two numbers b and c is a and the two G.M.s between them are g1 and g2. Ifg1

3 + g23 = k abc, then k is equal to

a) 1 b) 2 c) 3 d) 4

7. Between 1 and 31 are inserted m arithmetic means, so that the ratio of the 7th and (m – 1)th means is5 : 9. Then the value of m is

a) 12 b) 13 c) 14 d) 15


RAMKI

8. If the A.M. and G.M. of two numbers are 13 and 12 respectively then the two numbers are

a) 8, 12 b) 8, 18 c) 10, 18 d) 12, 18

9. Sum to n terms of the series

1 1 1 ..........

1 x 1 2x 1 2x 1 3x 1 3x 1 4x

is

(A) nx

1 x 1 nx (B) n

1 x 1 n 1 x

(C) x

1 x 1 n 1 x (D) none of these

10.n 1

1(n 1) (n 2)(n 3)....(n k)

is equal to

(A) 1

(k 1) k 1 (B) 1

k k

(C) 1

(k 1) k (D) 1k

11. Sum of the series

1 1 1S 1 1 2 1 2 3 1 2 3 4 .........2 3 4

upto 20 terms is

(A) 110 (B) 111

(C) 115 (D) 116

12. The sum of first n terms of the series

12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 + ........... is n (n + 1)2/2 when n is even. When n is odd the sum ofthe series is

(A) n2 (3n + 1)/4 (B) 2 n 1

n2

(C) n3 (n – 1)/2 (D) none of these


RAMKI

13. Sum of the series n

r 1

rr 1 ! is

(A) 11n!

(B) 11

n 1 !

(C) 12

n 1 !

(D) none of these

14. If 2p + 3q + 4r = 15, the maximum value of p3q5 r7 will be

(A) 2180 B) 4 5

15

5 .32

(C) 5 7

17

5 .72 .9

(D) 2285

15 If a1, a2, ... an are positive real numbers whose product is a fixed number c, then the minimum value of a1

+ a2 + ... an–1 + 2an is

(A) n(2c)1/n (B) (n + 1)c1/n

(C) 2nc1/n (D) (n + 1) (2c)1/n

More Than One Correct Answer Type Questions

16. If a1, a2, ............ an are in H.P. and d be the common difference of the corresponding A.P. then theexpression a1a2 + a2a3 + ......... + an–1an is equal to

(A) 1 na ad

(B) (n – 1) (a1 – an)

(C) n (a1 – an) (D) (n – 1)a1an

17.2n

n 0For 0 , if x cos ,

2

2n 2n 2n

n 0 n 0y sin , z cos sin , then

(A) xyz = xz + y (B) xyz = xy + z

(C) xyz = x + y + z (D) xyz = yz + x

18. If a, b, c are in H.P. , then the expression

1 1 1 1 1 1Eb c c c a b

equals

(A) 2

2 1bc b

(B) 2 2

1 3 2 14 cac a

(C) 2

3 2abb

(D) none of these


RAMKI

19. Three positive numbers form a GP. If the middle number is increased by 8, the three numbers form an AP.If the last number is also increased by 64 along with the previous increase in the middle number, theresulting numbers form a GP again. Then

(A) common ratio = 3 (B) first number = 4/9

(C) common ratio = –5 (D) first number = 4

20. Let the harmonic mean and the geometric mean of two positive numbers be in the ratio4 : 5. Then the two numbers are in the ratio

(A) 1 : 4 (B) 4 : 1

(C) 3 : 4 (D) 4 : 3

21. Between two unequal numbers, if a1, a2 are two AMs; g1, g2 are two GMs and h1, h2 are two HMs theng1.g2 is equal to

(A) a1h1 (B) a1h2

(C) a2h2 (D) a2h1

22. If a1, a2, a3, ............... are in A.P. with common difference d, then tann

1

r r 1r 1

d1 a a

equals

(A)

tan 1 n 1 1

n 1 1

a a1 a a (B)

tan 1

n 1 1

nd1 a a

(C) tan–1 an+1 – tan–1 (a1) (D) p/ 2

23. Let a1, a2, a3, ........ be terms of an A.P.

If 2

1 2 p 62

1 2 q 21

a a ............. a ap ,p q thena a .......... a aq

must be

(A) less than 1 (B) 27

(C) 1141

(D) 72

24. The sum of the series 1 + 2.2 + 3.22 + 4.23 + 4.23 + .......... + 100. 299

(A) is more than 2106 (B) is 99.2100 + 1

(C) is 100.2100 + 1 (D) is 99.299 + 1

25. Let the sum of the series 3 3 3 3 3 3

1 1 2 1 2 ......... n...........1 1 2 1 2 .......... n

upto n terms be

Sn ,n = 1, 2, 3, ........... Then Sn cannot be greater than

(A) 1/2 (B) 1 (C) 2 (D) 4


RAMKI

(Assertion – Reason Type Questons )Each question contains STATEMENT – 1 (Assertion) and STATEMENT – 2 (Reason). Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

26. Let a1, a2, a3, ........... be a sequence of real numbers such that sum to n terms of the sequence is

n 1n 2 2 n N

STATEMENT-1 : If tr = ar/2r, then t1, t2, t3 ........... forms an A.P.

STATEMENT-2 : If Tr = r/ar, then T1, T2, T3, ........... forms a G.P.

27. STATEMENT-1 : If a, b, c are three positive real numbers such that a c b and

1 1 1 1 0a a b c c b

then a, b, c are in H.P..

because

STATEMENT-2 : If a, b, c are distinct positive real numbers such that

a (b – c)x2 + b (c – a) xy + c (a – b)y2 is a perfect square, then a, b, c are in H.P.

28. Let a, b, c be three positive real numbers which are in H.P.

STATEMENT-1 : a b c b 42a b 2c b

because

STATEMENT-2 : If x > 0, then 1x 4x

Comprehension Type QuestionsThis section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered.Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.P29–31 : Paragraph for Question Nos. 29 to 31

Suppose x1, x2 be the roots of ax2 + bx + c = 0 and x3, x4 be the roots of px2 + qx + r = 0.

29. If 1 23 4

1 1x , x , ,x x are in A.P. , then

2

2

b 4acq 4pr

equals

(A) a2/r2 (B) b2/q2

(C) c2/p2 (D) a2/p2

30. If a, b, c are in G.P. as well as x1, x2, x3, x4 are in G.P. then p, q, r are in

(A) A.P. (B) G.P.

(C) H.P. (D) A.G.P.

31. If x1, x2, x3, x4 are in G.P., then its common ratio is

(A) 1/ 4

arcp

(B) 1/ 3

crap

(C) crap (D)

apbq


RAMKI

P32–34 : Paragraph for Question Nos. 32 to 34

For k,n N , we define

B (k, n) = 1.2.3 ...... k + 2.3.4 ...... (k + 1) + ........ + n (n + 1) ........ (n + k – 1), S0 (n) = n

and Sk (n) = 1k + 2k + ...... + nk

To obtain value of B (k, n), we rewrite B (k, n) as follows:

k k 1 k 2 n k 1B k,n k! .........

k k k k

n n 1 ........ n kn k

k!k 1 k 1

32. S3(n) + 3S2(n) equals

(A) B (3, n) (B) B (3, n) – 2B (2, n)

(C) B (3, n) – 2B (1, n) (D) B (3, n) + 2B (1, n)

33. k k 1 1 0k 1 k 1 k 1 k 1

S n S n ........... S n S n1 2 k k 1

equals

(A) (n + 1)k (B) (n + 1)k – 1

(C) nk – (n – 1)k (D) (n + 1)k – (n – 1)k

34. n

2

k 1k k 1 k 2

(A) 1

20n (n + 1) (n + 2) (n + 3) (4n + 15)

(B) 1

20n (n + 1) (n + 2) (n + 3) (2n + 12)

(C) 1

20n (n + 1) (n + 2) (n + 3) (2n + 13)

(D) 1

20n (n + 1) (n + 2) (n + 3) (n + 14)


This section contains 2questions. Each question contains statements given in two column which have to bematched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. Theanswers to these questions have to be appropriately bubbles as illustrated in the following example.If the correctmatches are A–p, A–s, B–q, B–r, C–p, C–q and D–s, then the correctly bubbled 4 × 4 matrix should be as follows:

____________________________________________________________________________________________


RAMKI

35. Match the conditions for the equation ax3 + bx2 + cx + d = 0 having roots in A.P.

Column I Column II

(A) AP (P) b3d = ac3

(B) GP (Q) 27ad3 = abcd2 – 2c3 d

(C) HP (R) 2b3 – 9abc + 27a2d = 0

(D) 3a = b +g (S) 4ad -bc = 0

(T) ad + bc

36. If n > 1 Sum of all the terms in the nth row of the triangle

Column I Column II

(A) 1 (p) n3

2 3

4 5 6

7 8 9 10

(B) 1 (q) 21 n n 12

1 1

1 2 1

1 3 3 1

(C) 1 (r) 2n–1

3 5

7 9 11A

13 15 17 19

(D) 12 (s) 2n–2Cn–2

12 12

12 22 12

12 32 32 12

Practice Questions


37. If the pth, qth and rth terms of an A.P. be a, b,c respectively, then

a q r b r p c p q

(A) 0 (B) 2

(C) p + q + r (D) pqr


RAMKI

38. If n n n

n 0 n 0 n 0x a ,y b ,z c

where a, b, c are in A.P. such that |a| <1, |b| < 1and

\|c| < 1, then x, y, z are in

(A) A.P. (B) G.P.

(C) H.P. (D) None of these

39. log3 2, log6 2, log122 are in

(A) A.P. (B) G.P.

(C) H.P. (D) none

40. In a G.P. if the (m + n)th term be p and (m – n)th term be q, then its mth term is

(A) pq (B) p / q

(C) q /p (D) p/q

41. If the roots of the equation

x3 – 12x2 + 39x – 28 = 0 are in A.P., then their common difference will be

(A) 1 (B) 2

(C) 3 (D) 4

42. If a, b, c, d are nonzero real numbers such that (a2 + b2 + c2) (b2 + c2 + d2) (ab + bc + cd)2, then a, b, c,d are in

(A) AP (B) GP

(C) HP (D) none of these

43. Value of 0.25 2 31 1 1log ......... upto3 3 3y 0.36

is

(A) 0.9 (B) 0.8

(C) 0.6 (D) 0.25

44. Let a1, a2, a3, ... be in AP and ap, aq, ar be in GP. Then aq : ap is equal to

(A) r pq p (B)

q pr q

(C) r qq p (D) none of these

45. Three distinct real numbers a, b, c are in G.P. such that a + b + c = x b, then

(A) 0 < x < 1 (B) –1 < x < 3

(C) x < –1 or x > 3 (D) –1 < x < 2


RAMKI

46. If x = 1 + y + y2 + y3 +..... to , then y is

(A) x

x 1 (B) x

1 x

(C) x 1

x

(D) 1 x

x

47. If exp. {(sin2 x + sin4 x + sin6 x + .... in f) loge 2} satisfies the equation x2 – 9x + 8 = 0, t hen the value of

cos x ,0 xcos x sinx 2

is

(A) 1 3 12

(B) 1 3 12


48. If x, y, z are in G.P., ax = by = cz, then

(A) logc b = loga c (B) log1 c = logb c

(C) loga b=logc b (D) logb a = logc b

49. The rth, sth and tth terms of a certain G.P. are R, S and T respectively, then the value of

Rs –t. S

t–r. Tr – s is

(A) 0 (B) 1

(C) –1 (D) none of these

50. The greatest value of x2y3z4, (if x + y + z = 1, x, y, z > 0) is

(A) 9

5

23

(B) 10

15

23

(C) 15

10

23

(D) none of these

51. If a, b and c are three positive real numbers, then the minimum value of the expression

b c c a a ba b c

is

(A) 1 (B) 2

(C) 3 (D) 6More Than One Correct Answer Type Questions

52. If n n n

n 0 n 0 n 0x a , y b , z c ,

where a, b, c are in A.P and |a| < 1, |b| < 1, |c| < 1 then x, y, z

(A) are all more than 1 (B) are all less than 1

(C) are in G.P. (D) are in H.P.


RAMKI

53. If 2 3

1 1 1S ........ upto3 3 3

, then

(A) 2log S0.25 4 (B) 5log S0.008 8

(C) 5log S0.008 4 (D) 2log S0.25 8

54. The sum of the first n terms of the series

22 2 2 2 2 2 n n 11 2.2 3 2.4 5 2.6 ..........is

2

,

when n is even. When n is odd, the sum is

(A) 2n n 12

(B) 2 2n 1 n

2

(C) even, if odd ‘n’ of the type 4 l + 1. (D) even, if the odd ‘n’ is of the type 4l + 3

55. The value of n

r 1

1a rx a (r 1)x

is

(A) n

a a nx (B) a nx a

x

(C) n( a nx a)

x

(D) none of these

56. If positive numbers a, b, c, d are in harmonic progression and a b , then

(A) a + d > b + c is always true (B) a + b > c + d is always true

(C) a + c > b + d always true (D) ad > bc

(Assertion – Reason Type Questions )Each question contains STATEMENT – 1 (Assertion) and STATEMENT – 2 (Reason). Each question has 4choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

57. STATEMENT-1 : There exists no A.P. whose three terms are 3, 5 and 7

because

STATEMENT-2 : If tp, tq and tr are three distinct terms of an

A.P., then r p

q p

t tt t

is a rational number


RAMKI

58. STATEMENT-1 : There exists an A.P. whose three terms are 2, 3, 5

STATEMENT-2 : There exists distinct real number p, q, r satisfying

2 A p 1 d, 3 A q 1 d

5 A r 1 d

59. STATEMENT-1 : The sum of an infinite A.G.P.

a + (a + d) x + (a + 2d) x2 + (a + 3d) x3 + ........... where |x| < 1 always exist

STATEMENT-2 : The sum of the infinite series a + ar + ar2 + ............. converges if |r| < 1.Comprehension Type QuestionsThis section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered.Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.P60–62 : Paragraph for Question Nos. 60 to 62

Sum of the following three series is given

1 1 1 11 ............. log22 3 4 5

(1)

1 1 1 1 11 .............3 5 7 9 11 2 2

(2)

1 1 1 11 .............3 5 7 9 4

(3)

60. 1 1 11 2 .............

3 5 7 9 11 13

is

(A) 2

(B) 12

(C) 4

(D) 14

61. Sum of series

1 1 1 11 ............upto7 9 15 17

is

(A) 2

(B) 1 28

(C) 14 2

(D) 4 2


RAMKI

62. Sum of the series

1 1 1 1 1 11 .......... upto2 3 4 5 6 7

is

(A) log 24 (B) log 2

4

(C) p – log 2 (D) p + log 2


This section contains 2questions. Each question contains statements


in Column I have to be matched with statements (p, q, r, s) in Column II.

The answers to these questions have to be appropriately bubbles



then the correctly bubbled 4 × 4 matrix should be as follows :

63. Let a, b, c, p > 1 and q > 0. Suppose a, b, c are in G.P.

Column I Column II

(A) logp a, logp b, logp c are in (p) G.P.

(B) loga p, logb p, logc p are in (q) A.G.P.

(C) a logp c, b logp b, c logp a (r) H.P.

(D) qloga

ap, qlogb b

p, qlogc cp are in (s) A.P.

64. Match the following, if f(n) denotes the sum of the series12 + 2.22 + 32 + 2.42 + 52 + 2.62 + …. Upto n terms, thenColumn I Column II

(A) f(49) (p) 21 (49) .50

2(B) f(50) (q) 25(51)2

(C) f(51) (r) 21 (51) 52

2(D) f(52) (s) 26(53)2

(t) 2.1 50 492


RAMKI

LEVEL-II

(ANSWER KEY)

1-C 2-C 3-A 4-B 5-A 6-B

7-C 8-B 9-B 10-C 11-C 12-B

13-B 14-C 15-A 16-A,D 17-B,C 18-A,B,C19-A,D 20-A,B 21-B,D 22-A,B,C 23-A,B 24-A,B

25-C,D 26-C 27-C 28-C 29-A 30-B

31-A 32-C 33-B 34-A

35. A P BQ C R DS

36 A Q BR CP DS

37-A 38-C 39-C 40-A 41-C 42-B

43-C 44-C 45-C 46-C 47-B 48-D

49-B 50-B 51-D 52-A,D 53-A,B 54-A,D

55-A,B 56-A,D 57-A,B 58-B 59-A 60-C

61-B 62-A 63. A S BR C QDP 64- A P BQ CR DS

HINTS AND SOLUTIONS

LEVEL-II

1. 3ar 1 r 216 and 3

5

ar 14ar

2r 4 r 2, 2

when r = 2 then 2a (9) = 216 a 12

when r = – 2, then – 2a (1 – 8) = 216

216 108a14 7

, which is not an integer..


RAMKI

2. 2 1

1 2 1 2 1 2

a a1 1 da a a a a a

1 2 1 2

1 1 1 1a a d a a

2 3 2 3

1 1 1 1a a d a a

...................................

...................................

n 1 n n 1 n

1 1 1 1a a d a a

whence by addition, we get

n 1

1 n 1 n

a a1 1 1 1Sd a a d a a

1 n 1 n

n 1 d n 11d a a a a

3. . 2

m2

n

S mS n

m n2 2

S S km n

(say)

22

m m m 122

n n n 1

k m m 1T S ST S S k n n 1

= 2m 12n 1

4.

n

n

n / 2 [2a n 1 d]SS n / 2 2a n 1 d

= 3n 87n 15

or,

n 1a d2

n 1a d2

= 3n 87n 15

...(i)

We have to find 12

12

T a 11dT a 11d


RAMKI

Choosing (n – 1) / 2 = 11 or n = 23 in (i)

we get, 12

12

T 7T 16

5. Let numbers are a and b.

given that a b 5 a b 10

2

&

ab ab 16

and 2ab 2 16 16H.M.a b 10 5

6. b, a, c are in A.P. 2a b c

and b, g1, g2, c are in G.P.

2 21 2 1 2

1 2 2 1

g g g gc or b, cb g g g g

& g1g2 = bc

k 2

7. 1 7d 5 31 1& d

1 m 1 d 9 m 1

on solving we get m = 14

8. Let numbers are a and b

given that a b 13 & ab 12

2

solving these we get numbers 8, 18

9. If tr denotes the nth term of the series, then

rx 1 1xt

1 rx 1 r 1 x1 rx 1 r 1 x

n n

rr 1 r 1

1 1x t1 rx 1 r 1 x

1 1 nx

1 x 1 n 1 x 1 x 1 n 1 x

n

rr 1

nt1 x 1 n 1 x


RAMKI

10. n1T

n 1 n 2 n 3 ...... n k

n 11v

n 1 n 2 n 3 ..... n k 2 n k 1

n n 11 1 1v v

n 2 n 3 .... n k 1 n k n 1

1 k

n 1 n 2 n 3 ...... n k 1 n k

n n 1 nv v 1 k T

n n n 11 k T v v

1 1 01 k T v v

2 2 11 k T v v

3 3 21 k T v v

...............................

n n n 11 k T v v

_________________

Adding (1 – k) Sn = n 0v v

n1 1or 1 k s

n 1 n 2 ....... n k 1.2.3......k

Ltnn

11 k S 0|k

Ltn n

1Sk 1 |k

11. We have

k k 11 1 k 11 2 3 ........ kk k 2 2

Thus, 1 10S 2 3 4 .........21 2 21 1152 2


RAMKI

12. Let n = 2m, then

S2m = 12 + 2 × 22 + 32 + 2 × 42 + ............ + (2m – 1)2 + 2 (2m)2

= 2m (2m + 1)2/2 = m (2m + 1)

When n = 2m – 1

12 + 2 × 22 + 32 + 2 × 42 + ........... + (2m –1)2

= S2m – 2 (2m)2 = m (2m + 1)2 – 2 (2m)2

= m [4m2 + 4m + 1 – 8m] = m (2m – 1)2

= n2 (n + 1)/2

13. Let tr denote the rth term of the series, then

rr r 1 1 1 1t

r 1 ! r 1 ! r ! r 1 !

n n

rr 1 r 1

1 1 1t 1r ! r 1 ! n 1 !

14. 2p 3q 4r 15 ........ (i) given

By weighted mean

2p 3q 4r3 5 73 5 7

15

3 5 7

152p 3q 4r3 5 7

3 5 73 5 7

3 5 7

2 3 4or p q r 13 5 7

3 5 73 5 7

3 5 7

3 5 7or p q r2 3 4

5 73 5 7

17

5 7or p q r9.2

\ Maximum value of 5 7

3 5 717

5 7p q r is9.2

15 1 2 n 1 na a ........a .a c given

A.M. G. M.

or, 1 2 n 1 n n1 2 n

a a ...... a 2a a a ...... 2an


RAMKI

or, n1 2 n 1 na a ...... a 2a 2cn

n1 2 n 1 na a ..... a 2a n 2c

minimum value of 1 2 n 1 na a .... a 2a is nn 2c

16.1 2 1 2 1 2

1 22 1

a a d a a a a1 1a ad d a a d

Similarly 2 3 3 42 3 3 4

a a a aa a ,a a

d d

and so on

a1a2 + a2a3 + a3a4 + ........ + an–1an

1 2 3 4 n 1 n1 a a a a ............. a ad

1 na ad

...... (i)

But n 1

n 1

1 1a a1 1 n 1 d d

a a n 1

Equation (i) becomes (n – 1) a1an

17. 2 2x cosec , y sec , 2 2

1z1 cos sin

2 2

1xy andsin cos

2 2 2 2

1 1 1x ysin cos sin cos

x y xy

2 2 2 2

1 1 1x y zsin cos 1 sin cos

2 2 2 2

1 1 xyzsin cos 1 sin cos


RAMKI

18. As a, b, c are in H.P., 1/a, 1/b, 1/c are in A.P.

Let the common difference of this A.P. be d Now,

22

1 1 1E d d dc c c

2

2 2

1 1 1 2 1c b bcc b

Next, 2

2 2 2

1 1 1 1 1 3 2 1E4 c a 4 cac c a

and 2 2

2

2 1 1 1 3 2Eb a b a abb

19. Let a, ar, ar2 are in G.P., as per question

a, ar + 8, ar2 are in A.P.

\ a + ar2 = 2ar + 16 ....(i)

Also a, ar + 8, ar2 + 64 are in G.P.2ar 8 ar 64

a ar 8

ar + 4 = 4a

4a4 r

....(ii)

using (i) and (ii)

24 1 r 8r 164 r 4 r

1

4r2 + 8r – 60 = 0

r = 3, –5, r – 5

r 3

from equation (ii)

a = 4

20. On solving 2ab

4a b5ab

as a quadratic in ab ,

we get ab = 4, 1/4

a and b are the correct choices.


RAMKI

21. Let A, 1 2a ,a , B be in A. P..

1B A 2A Ba A

3 3

2B A A 2Ba A 2.

3 3

Also A, g1, g2, B are in G.P.

3B rA

1/31g Ar A B / A

2/322g Ar A B / A

21 2g g A B / A AB

Also A, h1, h2, B are in H.P.

1 2

1 1 1 1, , , are inAPA h h B

1

1 1 1 1 1 1 A Bh A 3 B A A 3AB

1

1 3B A B A 2Bh 3AB 3AB

13ABh

A 2B

and

2

3B 2 A B1 1 2 1 1h A 3 B A 3AB

2

1 2A Bh 3AB

23ABh

2A B

Obviously 1 2 1 2 2 1g g AB a h a h

22. Write d = a2 – a1 = a3 – a2 = ..........

and use 1 1 1x ytan tan x tan y

1 xy


RAMKI

23.

121

2

1 1

p 1p a d2a p 1 d p p22q q 1 qq2a q 1 d a d2 2

For 6

21

a,put p 6,q 21

a

6

21

a 2 1a 7

24. S = 1 + 2.21 + 3.22 + 4.23 + .......... + 100. 299

2S = 1.21 + 2.22 + 3.23 + .......... + 99. 299 + 100 .2100

Subtracting

–S = 1 + 21 + 22 + 23 + ......... + 299 – 100.2100

–S = (2100 – 1) – 100. 2100

S = 99. 2100 + 1

As 99 > 26

\ S > 2106

25. r 3 3 3

1 2 3 ......... rT1 2 ........... r

2 1 12

r r 1 r r 1

n n

n rr 1 r 1

1 1 1S T 2 2 1r r 1 n 1

which can not be greater than 2

26. Ans: (C)

Sol: n n n 1a S S

n 1 nn 2 2 n 3 2

n n nn.2 2 2 n 1

stat. 1 r

rr r r

2 r 1at r 1

2 2

1 2 3 4t , t , t , t are in A.P..

stat. 2 r rr

r rTa 2 r 1

which is not in G.P..


RAMKI

27. Ans: (B)

Sol: Rewrite the given expression as

1 1 1 1 0a c b c a b

a c b a c b 0a c b c a b

2aca c b c b a ba c

For statement-2, put t = x/y, and write the expression as

y2 [a (b – c) t2 + b (c – a) t + c (a – b)]

t = 1 satisfies the expression within bracket. For perfect square other zero must be 1.

28. Ans: (C)

Sol:2ac b 2c b 2ab anda c a a c c a c

a b 1 b / a a 3c2a b 2 b / a 2a

and c b 1 b / c 3a c2c b 2 b / c 2c

Thus, a b c b 3 c a 31 1 2 42a b 2c b 2 a c 2

29 . 1 2x x b / a, 1 2x x c / a,

3 4x x q/p, 3 4x x r / p,

. 2 14 3

1 1x xx x

22 3 4

2 1 24 3

x xx x

x x

22 1

2 24 3 3 4

x x 1x x x x

22 1 1 2

2 23 4 3 4 3 4

x x 4x x 1x x 4x x x x

On putting values we get2 2

2 2

b 4ac aq 4pr r


RAMKI

30.4 32 4 2 1

1 3 1 3

x xx x x xx x x x

2 21 2 1 1 2

2 23 433 4

x x x x xx xxx x

on putting value, we get 2q pr

31. 2 32 1 3 1 4 1x x r ,x x r ,x x r '

21 2 1x x x r c / a ......... (i)

2 33 4 1x x x r r / p ........ (ii)

dividing (ii) by (i) 1/4

4 ra rarpc pc

= r¢

32 . S3 (n) +3S2 (n)= n n n

2

k 1 k 1 k 1k k 3 k k 1 k 2 2 k

= B (3, n) – 2B (1, n )

33. . k 1 k 1 k 11 k 2 k 1 k 1S n S n .... S n

+ k 1k 1 0S n

= n

k 1 k k 1 k 1 k 1 k 11 2 k k 1

r 1r r ..... r

n

k 1 k 1k 1

r 1[ r 1 r ] n 1 1

34. k(k + 1) (k + 2)2 = k (k + 1) (k + 2) (k + 3) – k (k + 1) (k + 2)

n

2

r 1k k 1 k 2 B 4,n B 3,n

1n n 1 n 2 n 3 n 45

1n n 1 n 2 n 34

1 n n 1 n 2 n 3 4n 1120


RAMKI

35..Ans: A-P, B-Q, C-R, D-SSol: Let the roots be in A.P. and let they be a – b, a, a + b

then (a – b) + a + (a + b) = –b/aa (a – b) + a (a + b) + a2 – b2 = c/aa (a – b) (a + b) = d/a

From first relation b3a

From second relation 2 2 c3a

and from third relation

2 2 da

Since b3a

we can easily eliminate b2 in last two relation to get

2b3 – 9abc + 27a2d = 0We can similarly arrive at

36.

Ans: A-Q, B-R, C-P, D-S

Sol: A) Last term of nth row

= 1 + 2 + 3 + ........ + n = 1/2 n (n + 1)

Sum of terms in the nth row

2n 1n n 1 n 1 1 n n 12 2

B) Its Pascal’s triangle

\ Sum to n terms = 2n–1

C) Last term of nth row

= mth odd number where m equals 1 n n 12

= n2 + n – 1

\ Sum of terms in the nth row

2 31 n 2n 2n 2 2 n 1 n2

D) Sum to nth row

2 2 2n 1 n 1 n 10 1 n 1C C .......... C

= 2n–2Cn–1.


RAMKI

37. TP = A + ( p – 1) D = a

or a = (A – D) + pD etc. for b and c

Multiply by q – r, r – p, p – q and add and use

sigma i.e. a b 0, a b c 0

38. 1 1x ,y

1 a 1 b

z =

11 c

1 1 1a 1 ,b 1 ,c 1x y z

1 1 11 ,1 ,1x y z

are in AP

x, y, z are in H.P..

39. If the numbers be x, y, z then

1/x = log23,1/y = log2 2.3 = 1 + log23,

1/z = log2 (4 × 3) = 2 + log2 3 which are in A.P.

x, y, z are in H.P..

40. m n 1m nT ar p

m n 1m nT ar q

Multiplying 2 2m 2a r pq

m 1mT ar pq

41. Sum of three numbers in A.P.

= 3a = 12

2x 4 x 8x 7 0

x 1,4,7 or 7,4,1, d 3

42. (a2 + b2 + c2) (b2 + c2 + d2) (ab + bc + cd)2,

Solving this we get

4 4 2 2 2 2 2 2 2 2 2 2b c a c a b b d b c 2ab c 2bc d 2abcd 0

or 2 2 22 2b ac c bd ad bc 0


RAMKI

\ 2 b cb ac 0a b

, 2 c dc bd 0b c

, d bab bc 0c a

\ b c da b c Hence a, b, c, d are in G.P..

43. We have

2 31 1 1 1/ 3 1S ..........3 1 1/ 3 23 3

1/ 2

0.25logy 0.36

log 1/ 2logy log 0.36

log 0.25

log2 1log 0.36 log 0.36log4 2

y 0.6

44. Let c.d = d

p 1a a p 1 d , q 1a a q 1 d , r 1a a r 1 d

as p q ra ,a ,a are in G.P..

q q rr

p q p q

a a aaa a a a

(by law of proportions)

or

q 1 1r

p q 1 1

a a q 1 d a r 1 daa a a p 1 d a q 1 d

=q rp q or

q

p

a q r r qa p q q p

45. a + ar + ar2 = x. ar

or, r2 + r (1 – x) + 1 = 0, r is real

20 i.e. 1 x 4 0

or, x2 – 2x – 3 > 0

or, (x + 1) (x – 3) > 0

x < – 1 or x > 3


RAMKI

46.1x

1 y

47.2 2

22 2

sin x sin xS tan x1 sin x cos x

22 tan 2tan xlog2 log2 x tan xLHS e e 2 .... (i)

and the roots of x2 – 9 x + 8 = 0 are 1 and 8

2 2tan x 0 tan x 32 1 2 ,2 8 2

2 2tan x 0, tan x 3 or tan x 0, tan x 3

x3

is the only value of x s.t. 0 < x < 2

cos x 1 1cos x sin x 1 tan x 1 3

3 1 1 3 13 1 2

48. x log a = y log b = z log c = k (say)

Also y2 = xz

2 2

2

k kloga.logclogb

or loga logblogb logc

or logb a = logc b

49. Let the common ratio be taken as k and a be the first term.

r 1rR T ak

r 1 s ts t s tR a k similarly

s 1 t rt r t rS a k

t 1 r sr s r sT a k

Multiplying the above three and knowing that

m n p m n pA .A .A A

s t 1 r r s 0 0R S T a .k 1


RAMKI

50.x y zx y z 1 2. 3. 4. 12 3 4

using weighted mean

12 3 4 9

x y z2. 3. 4. x y z2 3 42 3 49

9 2 3 4 102 3 4

10 3 15

1 x y z 2x y z9 2 3 3

51. AM GM

b c c a a ba a b b c c

6

1/ 61

minimum value is 6

52.1 1 1x ,y ,z

1 a 1 b 1 c

as a, b, c are in A.P..

1 – a, 1 – b, 1 – c will be in A.P.

1 1 1. ,1 a 1 b 1 c

will be in H.P..

x, y, z will be in H.P.

Choice (d) is correct

As – 1 < a < 1 1 a 1 x 1

Similiarly , y > 1, z > 1

choice a is also correct.

53. We have 1/ 3 1S

1 1/ 3 2

Thus, 2 2log S 2log 1/20.25 0.5

= (0.5)–2 = 4

5 5log S 3log 1/20.008 0.2

53log 2

1 85


RAMKI

54. If n is odd then n – 1 is even and Sn = Sn-1 + n2

2 2

2n 1 n nn n 12 2

since 2n n 1

2 is even if n is of the from 4 I + 3 .

55.n

r 1

1a rx a (r 1)x

=

n

r 1

a rx a r 1 x

a rx a r 1 x

= n

r 1

1 a rx a r 1 xx

= 1 a x a a 2x a x .... a nx a n 1 xx

= a nx a

x

=

a nx a nx a nx a a nx a

56. Take 1 1 1 1a , b ,c ,d

p 3q p q p q p 3q

Then a + d > b + c easily follows

Since (a + d) – (b + c)

2

2 2 2 2 2 2 2 2

2p 2p 8q2pp 9q p q p 9q p q

Which is positive a, b,c,d 0

Also 2 2 2 2

1 1ad bcp 9q p q

2

2 2 2 2

8q 0p 9q p q


RAMKI

57. Ans: (A)

Suppose 3, 5 and 7 are the pth, qth and rth terms of an A.P. whose common difference is d, then

tr – tp = (r – p) d

and tq – tp = (q – p) d

r p

q p

t t r pt t q p

which is rational numbers

7 35 3

is a rational numbers.

7 3 5 3

5 3

is rational

35 15 21 3

2

is rational

35 15 21 is rational, say r..

Now,

35 15 21 r

15 21 35 r

Squaring both sides, we get

215 21 2 6 35 35 r 2r 35

21 r3512 2r

35 is rational

This is a contradiction

Hence 3, 5 and 7 cannot be three terms of an A.P..

58. Sol. (B)If we could show that Reason R is false, then Assertion A will also be false. Indeed if R istrue, then

,2 3 p q d 3 5 q r d

On dividing 2 3 p q

q r3 5

Rational = IrrationalBoth A and R are false


RAMKI

59. Ans: (A)

Sol.S = a + (a + d) x + (a + 2d) x2 + ..........xS = ax + (a + d) x2 + (a + 2d) x3

S (1 – x) = a + dx + dx2 + dx3 + ........ = a + dx

1 xinfinite GP with common ratio xSince |x| < 1

2a dxS

1 x (1 x)

S is infinite and therefore converges.60.

Sol: The given series can be written as

1 1 1 1 1 11 .........3 5 7 9 11 13 4

61. Add series (2) and (3)

62. Multiply (1) by 1/2 and subtract from (3.)

63. A-S, B-R, C-Q, D-P

Sol. Since b2 = ac

2log b = log a + log c

log a, log b, log c are in A. P..

(a) p p plog a,log b,log c in A.P..

(b) Since p p p

1 1 1, ,log a log b log c in H.P..

a b clog p,log p,log p in H.P..

(c) since a, b, c in G.P.

and p p plog c,log b,log a in A.P..

p p palog c,blog c,c log a

(d) on checking given number are in G.P.

64. Case (i) : when n is an evenLet n = 2m\ f(2m) = (12 + 2.22) + (32 + 2.42) + ........... + m termsTm = (2m – 1)2 + 2 [2+ (m – 1)2]2

Tm = 4m2 – 4m + 1 + 8m2 = 12m2 – 4m+1


RAMKI

2f 2m 12 m 4 m 1

= m m 1 2m 1 m m 1

12 4 m6 2

or, f (2m) = m (2m + 1)2

\ f (n) = 2n n 12

case (ii) when n is an odd\ (n – 1) is even

So 2 22n 1 n n n 1

f n n2 2

2n n 1f n

2

A) 2

249 49 1 1f 49 49 .502 2

B) 2

250 50 1f 50 25. 51

2

C) 2

251 51 1 1f 51 51 .522 2

D) 2

252 52 1f 52 26. 53

2

LEVEL-III

MODEL QUESTIONS


1. If a, a1, a2, a3, ... a2n–1, b are in AP, a, b1 , b2 , b3 ..., b2n–1, b are in GP and a, c1, c2, c3, ... , c2n–1, b are in HP,where a, b are positive, then the equation anx

2 – bnx + cn = 0 has its roots

(A) real and unequal (B) real and equal

(C) imaginary (D) none of these

2. The harmonic mean of two numbers is 4. Their arithmetic mean is A and geometric mean is G. If G satisfies2A + G2 = 27, the numbers are :

a) 1, 13 b) 9, 12 c) 3, 6 d) 4, 8

3. For a positive integer n, let

n

1 1 1 1a n 1 ......2 3 4 2 1

. Then

(A) a (100) < 100 (B) a (100) > 100

(C) a (200) < 100 (D) none of these


RAMKI

4.r 1n

r 1

1r 1n

is equal to

(A) 2n2 (B) 3n2

(C) n2 (D) none of these

5. Let n n

44

r 1 r 1r f n , then 2r 1

is equal to

(A) f (2n) – 16 f (n) (B) f (2n) – 7f (n)

(C) f (2n – 1) – 8 f (n) (D) none of these

6. If n 1 3 1n

1a for n 1 and a a ,1 a then (a2001)2001 equals

(A) –1 (B) 1


7. If n n

rrr 1 r 1

1 1t n n 1 n 2 , then value is12 t

(A) 2n

n 1(B)

n 1n 1 !

(C) 4n

n 1 (D) 3n

n 2

8. If n n

2n r r

r 1 r 1

1S t n 2n 9n 13 , then t6

equals

(A) 1n n 12

(B) 1 n n 22

(C) 1 n n 32

(D) 1 n n 52

9. If 2

2 2 21 1 1 ......upto

61 2 3

, then value of

2 2 21 1 1 .......... upto to is1 3 5


RAMKI

(A) 2

4

(B) 2

6

(C) 2

8

(D) 2

12


1 1 1 .......... is1.2.3.4 2.3.4.5 3.4.5.6

(A) 1 1

24 n (B)

1 118 3 n 1 n 2 n 3

(C) n 21 12 n

(D) 1 1

18 3 n 3 n 4

11. Sum to n term of the series

21 1S 1 2 1 3 1 ........ is given byn n

(A) n2 (B) (n + 1)2

(C) n (n + 1) (D) none of these

12. Sum to n terms of the series 1 1! 2! 3! .......... is5! 6! 7! 8!

(A) 2 15! n 1 !

(B)

1 1 n!4 4! n 4 !

(C) 1 1 3!4 3! n 2 !

(D) none of these


1 3 5 7 ....1.2.3 2.3.4 3.4.5 4.5.6

is

(A)

n n 1

2 n 2 n 3

(B)

n 3n 1

4 n 1 n 2

(C) 1 56 n 1 n 4

(D) none of these


RAMKI

14. Let p,q,r R and 327pqr p q r and 3p 4q 5r 12 then

p3 + q4 + r5 is equal to

(A) 3 (B) 6


15. In an acute angled triangle DABC, the minimum value of tann A + tann B + tann C. is

(when n N, n 1 )

(A ) n23 (B) 3n

(C) n 123 (D)

n 123

More than One Correct Answer Type Questions

16. If the numbers a b b c, b and1 ab 1 bc

are in A.P., then

(A) a, b, c are in HP (B) a, b–1, c are in HP

(C) a–1, b, c–1 are in AP (D) a, b, c are in GP

17. The three sides of a right-angled triangle are in G.P.. The tangents of the two acute angles may be

(A) 5 1 5 1and2 2

(B) 5 1

2

(C) 15 and5

(D) 5 1

2

18. If a, b, c are in H.P. then

(A) a b c, ,

b c a c a b a b c are in H.P.. (B)

2 1 1b b a b c

(C) b b ba , ,c2 2 2

are in G.P.. (D) a b c, ,

b c c a a b are in H.P..

19. If a, b, c are in A.P., and a2, b2, c2 are in H.P., then

(A) a = b = c (B) 2

2 2 ca b2

(C) a, b, c are in G.P. (D) a ,b,c are in G.P.

2


RAMKI

20. If (m + 1)th, (n + 1)th and (r + 1)th terms of an A.P. are in G.P. and m, n, r are in H.P., then the ratio of the firstterm of the A.P. to its common difference is

(A) n2

(B) m2

(C) r (D) mr

m r

21. If p, q, r are positive and are in A.P. the roots of the quadratic equation px2 + qx + r = 0 are all real for

(A) r 7 4 3p (B)

q 4 2 3p

(C) p 4 4 3q (D)

p 31 4q 2

22. If roots of x3 + bx2 + cx + d = 0 are

(A) in A.P. then 2b3 – 9bc + 27d = 0 (B) in G.P. then b3d = c3

(C) in G.P. then 27d3 = 9bcd2 – 4c3d (D) equal then c3 = b3 + 3bc

23. Let a and b be two positive real numbers. Suppose A1, A2 are two arithmetic means;G1, G2 are two geometric means and H1, H2 are two harmonic means between a and b then

(A) 1 2 1 2

1 2 1 2

G G A AH H H H

(B) 1 2

1 2

G G 5 2 a bH H 9 9 b a

(C) 1 2

1 2

H H 9abA A 2a b a 2b

(D) 1 2 1 2

1 2 1 2

G G H HH H A A

24 The value of 100 1 1 1 1......

1.2 2.3 3.4 99.100

(A) is an integer (B) lies between 50 and 98

(C) is 100 (D) 99

25. Let Sn = (1) (5) + (2) (52) + (3) (53) + ........... + (n) (5n) a1 4n 1 5 b16

, then

(A) a = n + 1 (B) a = n

(C) b = 5 (D) b = 25



RAMKI

26. Let a, b, c be three distinct non-zero real numbers

STATEMENT-1 : If a, b, c are in A.P. and b, c, a are in G.P., then c, a, b are in H.P.

STATEMENT-2 : If a, b, c are in A.P. and b, c, a are in G.P. than a : b : c = 4 : 2 : –1

27. STATEMENT-1 : 27, 8 and 12 can be three terms of G.P. as well as an A.P.

STATEMENT-2 : Three non-zero real numbers can always be three terms of an A.P.

28. STATEMENT-1 : If n is odd, 12 + (3) (22) + 32 + (3) (42) + 52 + ........ upto n terms

1 n n 1 4n 16

STATEMENT-2 : If n is even, 12 + (3) (22) + 32 + (3) (42) + 52 + ....... upto n terms

1 n n 1 4n 56

Comprehension Type QuestionsThis section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered.Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correP29–31 : Paragraph for Question Nos. 29 to 31

Let A1, G1, H1 denote the arithmetic, geometric and harmonic means, respectively of two distinct

positive numbers. For n > 2, let An – 1, n 1G and Hn – 1 has arithmetic, geometric and harmonic meansrespectively.

29. Which of the following statements is a correct statement ?

(A) 1 2 3G G G .... (B) 1 2 3G G G ....

(C) 1 2 3G G G .... (D) 1 3 5G G G .... and 2 4 6G G G ...

30. Which of the following statement is correct ?

(A) 1 2A A ....

(B) 1 2 3A A A ...

(C) 1 3 5A A A ... and 2 4 6A A A ...

(D) 1 3 5A A A ... and 2 4 6A A A ...

31. Which of the following is a correct statement ?

(A) 1 2 3H H H ...

(B) 1 2 3H H H ...

(C) 1 3 5H H H ... and 2 4 6H H H ...

(D) 1 3 5H H H ... and 2 4 6H H H ...


RAMKI


Consider the sum n n 4

8 16 8rS given by S ........5 65 4r 1

32. The value of limn nS must be

(A) 0 (B) 1/2 (C) 2 (d) None of these

33. The value of S16 must be

(A) 8041

(B) 1088545

(C) 107245

(D) None of these

34. If 2

n 3 2an bnS

cn dn en 1

where a, b, c, d, e are independent of n then

(A) a = 4, e = 2 (b) c = 0, d = 4

(C) b = 4, e = 4 (D) None of these


This section contains 2questions. Each question contains statements given in two column which have to bematched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. Theanswers to these questions have to be appropriately bubbles

as illustrated in the following example.If the correct matches are A–p, A–s, B–q, B–r, C–p, C–q and D–s, then thecorrectly bubbled 4 × 4 matrix should be as follows :

______________________________________________________________________________

35. Let y = exp {(sin2 x + sin4 x +sin6 x +.....)loge 2} Satisfy the equation. x2- 17x +16 = 0 and 0 < x < p/2

(A) 22sin2x

1 cos x(P)1

(B) 2sinx

sinx cos x(Q) 4/9

(C) n

n 1(cot x)

(R) 2/3

(D) 2n

n 1n(cot x)

(S) 4/3


RAMKI

36. Let a, b, g be three numbers such that 2 2 2

1 1 1 1 1 1 1 9,2 4

, and a + b + g = 2.

Column I Column II(A) a b g (p) 6(B) bg + ga + ab (q) 8(C) a2 + b2 + g2 (r) –2(D) a3 + b3 + g3 (s) –1

(t) even integer

Practice Questions

Single Correct Answer Type Questions37. If a, b, c are three unequal numbers such that a, b, c are in A.P. and b – a, c – b, a are in G.P.,, then a

: b : c is

(A) 1 : 2 : 3 (B) 1 : 3 : 5

(C) 2 : 3 : 4 (D) 1 : 2 : 4

38. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is11, the number of terms is

(A) 10 (B) 11


39. The numbers t 2 21t 1 , t2

and 6 are three consecutive tems of an A.P. If t be real, then the next two

terms of A.P. are

(A) –2, –10 (B) 14, 6

(C) 14, 22 (D) none

40. If a1, a2, a3, ...... an are in A.P., then the value of

1 2 2 3 n 1 n

1 1 1...a a a a a a

is

(A) 1 n

1a a (B)

1 n

1a a

(C) 1 n

na a (D)

1 n

n 1a a

41. If 1 2 na ,a ......a be an A.P. of positive terms, then

1 2n 2 2n 1 n n 1

1 2 2 3 n n 1

a a a a a a....a a a a a a


RAMKI

(A) 1 2n

1 n 1

n a aa a

(B) n – 1

(C) 1 n 1

n 1a a

(D) none

42. Three numbers form an increasing G.P. If the middle number is doubled, then the new numbers are inA.P. The common ratio of G.P. is

(A) 2 3 (B) 2 3

(C) 3 2 (D) none of these

43. Given than 0 x4

and y4 2 and k 2k

k 01 tan x p;

k 2k

k 01 cot y q;

then

2k 2k

k 0tan x cot y

is

(A) 1 1 1p q pq (B)

11 1 1p q pq

(C) p + q – pq (D) p + q + pq

44. If 1 2 2nA,A ,A ,....,A ,B be an A. P. ; A, 1 2 2nG ,G ,...,G ,B be a G.P. and H is the harmonic mean of A and

B, then

1 2n 2 2n 1 n n 1

1 2n 2 2n 1 n n 1

A A A A A A....G G G G G G

is equal to

(A) 2nH

(B) 2nH

(C) nH (D) nH

45. If H1, H2,....., Hn are n harmonic means between a and b( a), then the value of 1 n

1 n

H a H bH a H b

a) n + 1 b) n – 1

c) 2n d) 2n + 3

46. If the first and (2n+ 1)th terms of an A.P.; G.P. and H.P. are equal and their (n + 1)th terms area, b and c respectively, then

a) a > b > c b) ac = b2

c) a + b = c d) none of these


RAMKI


3 3 3 3 3 31 1 2 1 2 3 ...........is1 1 3 1 3 5

(A) 2n n 9n 1424

(B) 2n 2n 7n 1524

(C) 2n 2n 9n 1324

(D) 2n n 11n 1224

48. Sum of the series n

r 1

r 1r logr

(A) nn 1

logn!

(B) log n 1n!

(C) n!log n 1 (D) none of these

49. Sum of the series S = 12 – 22 + 32 – 42 + .......... – 20022 + 20032 is

(A) 2007006 (B) 1005004


50. If m be positive integer greater than 1 then

(A) 1m + 3m + 5m + ..... +(2n – 1)m > nm+1

(B) 1m + 3m + 5m + ......+(2n – 1)m > nm

(C) 1m + 3m + 5m + ......+(2n – 1)m > nm–1

(D) 1m + 3m + 5m + ......+(2n – 1)m > nm–2

51. The least value of the expression 5sinx–1 + 5-sinx-1 is

(A) 2/5 (B) 1/5

(C) 5 (D) 5/2More than One Correct Answer Type Questions52. Four numbers are such that the first three are in A.P., while the last three in G.P.. If the first number is 6

and common ratio of G.P. is 12

then the

(A) sum of first and last number is 7

(B) numbers are 6, 8, 4, 2

(C) numbers are 6, 10, 14, 4

(D) numbers are 6, 4, 2, 1


RAMKI

53. Let n

n times

a 111..........1 , then

(A) a912 is not prime (B) a951 is not prime

(C) a480 is not prime (D) a91 is not prime

54. x1, x2, x3, ............. is an infinite sequence of positive integers in G.P. such that x1x2x3x4 = 64. Thenthe value of x5 is

(A) is a perfect sequence (B) is not a perfect square

(C) 128 (D) 16

55. For a positive integer n, let n

1 1 1 1a n 1 ............2 3 4 2 1

; then

(A) a (100) < 100 (B) a (100) > 100

(C) a 200 100 (D) a (200) > 100

56. In the nth row of the triangle

1

2 3

4 5 6

7 8 9 10

(A) Last term 1n n 12

(B) Firs term 21 n n 22

(C) Sum 21n n 12

(D) Sum 21n n 12


57. STATEMENT-1 : If three positive numbers in G.P. represent sides of a triangle, then the common ratio

of the G.P. must lie between 5 1 5 1and2 2

STATEMENT-2 : Three positive real number can form sides of a triangle if sum of any two is greaterthan the third


RAMKI

58. Let a, b, c, d are four positive number

Statement–1 :a b c d a4b c d e e

Statement–2 :b c d e a 5a b c d e .

59. Let a, b, c and d be distinct positive real numbers in H.P.Statement–1 : a + d > b + c

Statement–2 :1 1 1 1a d b c

Comprehension Type QuestionsThis section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered.Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correP60–62 : Paragraph for Question Nos. 60 to 62

Let {an} be a sequence of real numbers. If limn na exists, we say that the sequence {an} is convergent.

An infinite series can also be defined with the help of a sequence. Indeed letS’ = b1 + b2 + b3 + b4 + ......... + bn + ................

Let us define a sequence {an} as a1 = b1 = S1, a2 = b1 + b2 = S2, a3 = b1 + b3

= S3, an = b1 + b2 + b3 + ........... + bn = Sn (sum of n term of series)

Infact {an} is a sequence of partial sums of the original series. We say that the series S’ is convergentprovided the sequence of partial sums is convergent. It can be noted that, if series b1 + b2 + b3 +

........... + bn + ........... of positive terms is convergent then limn nb 0 . Since

limn n n 1 n nb S S b 0 . Though the converse of this result is not true, it should be noted that

the convergence of a series is not affected if a finite number of terms are added or delected from theseries. It can also be obsereved that general series sum from the given series can be found by rear-rangement of terms

60. Let 2 2 2 2

1 1 1 1S ..............1 2 3 4

then 2 2 2 2

1 1 1 1 ...........1 3 5 7

is equal to

(A) 3 S8

(B) 3 S4

(C) 1 S4

(D) 1 S2

61. If S is same as given in above question then 2 2 2 2 2

1 1 1 1 1 ...........1 2 3 4 5

must be equal to

(A) S2

(B) S4

(C) 2S (D) None of these


RAMKI

62.1 1 1 1 11 .............2 3 4 199 200

must be equal to

(A) 1 1 1 1............1 2 3 100 (B)

1 1 12 1 ...........2 3 100

(C) 1 1 1 1............

101 102 103 200 (D)

1 1 1 1 1............101 102 103 104 200


This section contains 2questions. Each question contains statements given in two column which have to bematched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. Theanswers to these questions have to be appropriately bubbles

as illustrated in the following example.If the correct matches are A–p, A–s, B–q, B–r, C–p, C–q and D–s, then thecorrectly bubbled 4 × 4 matrix should be as follows :

_________________________________________________________________________________________________

63. Three positive numbers a, b, c satisfy

i) a + b + c = 25 ii) 2 < a, b, c < 18

iii) 2, a, b are in AP iv) b, c, 18 are in GP

then match the following

Column I Column II

(A) a (P) 12

(B) b (Q) 5

(C) c (R) 8

(D) 2a +3b +2c (S) 10

(T) odd integer

64. Match the following series with their summations

Column I Column II

(A) 12 – 32 + 52 – 72 + ........... – (4n – 1)2 (P) –8n2 – 4n

(B) 22 – 42 + 62 – 82 + ....... – (4n)2 (Q) –2n2 – n

(C) 12 – 22 + 32 – 42 + 52 ........... (4n)2 (R) –8n2

(D) 12 + 22 +32 +...........+(4n)2 (S) (8n2+ 2n)2


RAMKI

LEVEL-III

(ANS KEY)

1-C 2-C 3-A 4-C 5-A 6-A

7-C 8-C 9-C 10-B 11-A 12-B

13-B 14-A 15-B 16-B,C 17-B,D 18-A,B,C,D19-A,C,D 20-A,D 21-A,B,C,D 22-A,B 23-A,B,C 24-A,D

25-A,C 26-B 27-C 28-A 29-C 30-A31-B 32-C 33-B 34-A

35. A S BS C P DQ

36- A R,T BS CP,T DQ,T

37-A 38-B 39-C 40-D 41-A 42-B

43-B 44-A 45-C 46-B 47-C 48-A

49-A 50-B 51-A 52-A,D 53-A,B,C,D 54-A,D

55-A,D 56-A,B,C 57-A 58-A 59-B 60-B61-A 62-C

63- A Q,T BR CP DS

64. A R BP C P DS

HINTS AND SOLUTIONS

Level-III1. As odd number of AM, G.M and H.M. are inserted between a & b.

So, middle term of AP is AM = an

middle term of GP is GM = bn

middle term of HP is HM = cn

n n na ,b ,c are in G.P..

D = discriminant of quadratic equation < 0

roots are imagnary


RAMKI

2. Let numbers are a & b

Here a bA & G ab

2

given that 2A + G2 = 27 or (a + b) + ba = 27 ..... (i)

2abH.M. 4a b

..... (ii)

solving (i) & (ii), we get number 3 & 6.

3. We have

1 1 1 1 1 1a n 1 .......... ......... ......2 3 4 7 8 15

n 1

n 1 n 1 n 1n

1 1 1 2 4 8 2........ 1 ........ n2 4 82 2 1 22 1

Thus, a (100) < 100

4. r 1n

r 1

1r 1n

0 1 2 n 1

n1 1 1 1s 1 1 2 1 3 1 ...... n 1n n n n

2 n 1ns 1.1 2.x 3.x ..... nx .....(i) (where x = 1 +

1n

)

2 n 1 nnxs 1.x 2.x .... n 1 x nx .....(ii)

equation (i) – (ii), we get

n n

n 2

1 x nxs1 x1 x

n n

2

1 x nx 1 x

1 x

substituting 1x 1n

we get, 2ns n

5.

n n n

4 44

r 1 r 1 r 12r 1 r 2r

= f (2n) – 16f (n)


RAMKI

6. We have 1

2 31 2 1 1

1 a1 1 1a and a1 a 1 a 1 1/ 1 a a

Since a3 = a1, we get 21 1a 1 a

2 21 1 1a a 1 0 a w or w

We have

3

54 3 3

1 a1 1a1 a 1 1/ 1 a a

13 1

1

1 a a aa

Contrinuing in this way we obtain

a1 = a3 = a5 = a7 = a9 = a11 = ......... = a2001

Thus, (a2001)2001 = (– w)2001 or (–w2)2001 = –1

7. We have r r 1

r k kk 1 k 1

t t t

1 1r r 1 r 2 r 1 r r 112 12

1 r r 14

Now, r

1 4 1 14t r r 1 r r 1

n n

rr 1 r 1

1 1 1 1 4n4 4 1t r r 1 n 1 n 1

8. We have n n n 1t S S n 2

22n

1 1t n 2n 9n 13 n 1 2 n 1 9 n 1 136 6

3 23 21 2 n n 1 9 n n 1 13 n n 16

2 21 16n 6n 2 9 2n 1 13 6n 12n 66 6


RAMKI

= (n + 1)2

Also, t1 = S1 = 4 = (1 +1)2

n n

rr 1 r 1

1 1t r 1 n 1 n 2 1 n n 32 2

9. We have 2 2 21 1 1 ........ upto1 3 5

2 2 2 2 2 21 1 1 1 1 1 ........... upto1 2 3 4 5 6

2 2 21 1 11 ...........2 2 3

2 2 216 4 6 8

10. Here n1T

n n 1 n 2 n 3

n

n 3 nT

3n n 1 n 2 n 3

n1 1 1T3 n n 1 n 2 n 1 n 2 n 3

n n 11 V V3

n

n n 1 2 3 nn 1

S T T T T ..... T

n1 1 1S3 6 n 1 n 2 n 3

11. Let x = 1 + 1/n. Then

S = 1 + 2x + 3x2 ....... + nxn–1

2 n 1 nxS x 2x .......... n 1 x nx


RAMKI

Subtracting, we get

2 n 1 n1 x S 1 x x .........x nx

nn1 x nx

1 x

n n1 1 1S n 1 1 n 1 n

n n n

2S n

12. We have r r 1r 1 ! r !t and tr 4 ! r 5 !

Now, r r 1r ! r !rt r 5 t 0

r 4 ! r 4 !

n 1 n 1

r r 1 r 1 r 1 r r 1r 1 r 1

rt r 1 t 4t 4 t rt r 1 t

2 3 n 1 n4 t t ...... t 1t nt

1 2 n 1 n4 t t ........t 5t nt

n n 1 !0!55! n 4

1 n!4! n 4 !

1 2 n1 1 n!t t .........t4 4! n 4 !

13. We have r2r 1t

r r 1 r 2

=

2 1r 1 r 2 r r 1 r 2

=2 1 1 1 1 1

r 1 r 2 2 r r 1 r 1 r 2


RAMKI

Solving by using nv method

we get sum =

n 3n 1

4 n 1 n 2

14. 327pqr p q r

1/3 p q rpqr3

p q r

Also 3p + 4q + 5r = 12 p q r 1

15.nn n ntan A tan B tan C tan A tan B tan C

3 3

(Arithmetic mean of mth power of numbers)

nn n ntan A tan B tan C tan A tan B tan C3 3

sin ce tan A tan B tan c 3 3

n 1n n n 2tan A tan B tan C 3

16. The relation a b b c2b1 ab 1 bc

will yield a + c = 2 abc

17. Let the sides be a, ar, 2ar . If r > 1, then

(ar2)2 = (a)2 + (ar)2 (since in this case (ar)2 will be the hypotaneous i.e., the largest side)

4 2 2 1 5r 1 r r2

, 2 1 5r

2

is not possible

1 5r2

If 0 < r < 1 then a is the largest side

222 2a ar ar

5 1r

2


RAMKI

18.1 1 1, ,a b c

are in A.P..

a b c a b c a b c, ,a b c

are in A.P..

b c c a a b, ,a b c

are in A.P.. (Subtracting 1 from each term)

b c c a a b1, 1, 1a b c

are in A.P.. (Subtracting 1 from each term)

b c a c a b a b c, ,a b c

are in A.P..

Also 2acba c

so

2

2b a c 2b a c1 1b a b c b a b c b b a c ac

2

2 2

2b 2ac /b 2 b ac 2.b bb b 2ac b ac b ac

19. we have 2b = a + c and b2 = 2 2

2 2

2a ca c

.......(i)

On eliminating b, we get

2 2 2 2 2 28a c a c 2ac a c

which can be arranged as

2 2 2 2a c 2ac a c 4ac 0

either a = c or (a + c)2 + 2ac = 0

If a = c then a = b = c

a, b, c may be treated as three numbers in G.P..

If (a + c)2 + 2ac = 0, then by using (i) choice (d) follows.

20. Given (a + nd)2 = (a + md) (a + rd)

2a a an m rd d d

... (i)


RAMKI

Also n = m r n2mr mr

m r 2

... (ii)

Now from (i), 2 2

2a an a a2 n m r mrd d d d

22

m r nna n mr 2

d m r 2n m r 2n

from (ii)

a n mrd 2 m r

21. 2q = p + r, q2 – 4pr 0

Eliminate q to obtain p2 + r2 – 14 pr 0

Which gives (a) and (c)

Eliminate r to obtain

q2 – 8pq + 4p2 0

which gives (b) and (d)

22. Let roots be a – b, a, a + b, so that

/3 b b 3

3 3b b bc d 027 9 3

32b 9bc 27d 0

Next, roots be a/b, a, ab, so that

a3 = –d or a = (–d)1/3

\ – d + b (–d)2/3 + c (–d)1/3 + d = 0

3 3b d c

23. 1 21 2A a b a , A b b a3 3

1 2A A a b

Similarly, 1/ 3 2 / 3

1 2b bG a ,G aa a

1 2G G ab


RAMKI

and 21

1 1 1 1 1 1 2 1 1,HH a 3 b a a 3 b a

Now, 1 2

1 1 1 1H H a b

1 2 1 2

1 2 1 2

H H A Aa bH H ab G G

1 2 1 2

1 2 1 2

G G A AH H H H

Now, 1 23ab 3abH H

a 2b 2a b

9ab a ba 2b 2a b

2 21 2

1 2

2 a b 5abA AH H 9ab

Thus, 1 2

1 2

G G 5 2 a bA A 9 9 b a

24 The given series can be written as

2 1 3 2 100 99100 ......1.2 2.3 99.100

1 1 1 1 1 1100 .........1 2 2 3 99 100

= 99

25. 5Sn = (1) (5)2 + (2) (53) + .......... + (n – 1)5n + (n)5n+1

Subtracting from Sn, we obtain

–4Sn = 5 + 52 + .......... + 5n – n (5n+1)

nn 1

5 5 1n 5

4

n 1n

1S 4n 1 5 516


RAMKI

26. Ans: (B)

Sol: a + c = 2b, ab = c2 (i)

Now c a c2bc ac ab a

b c b c b c

\ a, b, c are in H.P.

Eliminating a from two expressions in (1) use get

2 2 2c / b c 2b c bc 2b 0

c b c 2b 0 c 2b c b

Thus, a = 4b.

Now, a : b : c = 4 : 1 : –2

27. Ans: (C)

Sol. Let if possible, 27 = arp–1, 8 = arq–1 and 12 = ars–1

,p q q s27 8r r8 12

q s p qp q q s q s p q27 8r and r

8 12

. .

q s p q 3q 3s p q27 8 3 2i e8 12 2 3

3q 3s q p or 2q p 3s For any p, q, s > 0, we get that 27, 8, 12 are three terms of G.P. A is correct and R is false

28. Ans: (A)

Sol: If n = 2m, then statement-2 becomes

12 + (3) (22) + 32 + (3) + (42) + 52 + ....... + 3 (2m)2 1 2m 2m 1 8m 56

2 22 2 2 21 3 2 3 3 4 5 ........ 3 2m 1

21 2m 2m 1 8m 5 3 2m6

2m 2m 1 8m 5 36m6

21 m 16m 18m 53


RAMKI

1 m 2m 1 8m 53

For n = 2m – 1, we get

12 + (3) (22) + 32 + (3) (42) + 52 +...... upto n terms

1 n 1 1n 4n 1 n n 1 4n 13 2 6

P29–31 : Paragraph for Question Nos. 29 to 31Sol. Let two distinct positive numbers be a and b.

1 1 11 2abA a b ,G ab,H2 a b

n n 1 n 1 n n 1 n 11A A H ,G A H2

n 1 n 1n

n 1 n 1

2A HHA H

For n > 2,

n n n 1 n 1A H A H

Thus, A1H1 = A2 H2 = A3H3 =....

2 2 21 2 3G G G ....

1 2 3G G G ....

As a and b are distinct, A1 > H1 1 2 2 1A A H H & 1 2 3 3 2 1A A A H H H

And so on.

Thus, 1 2 2 1A A H H

and 1 2 3H H H ....

P32–34 : Paragraph for Question Nos. 32 to 3432. c

n

n 4r 1

8rS4r 1

n

2 2r 1

8r2r 2r 1 2r 2r 1


RAMKI

n

2 2r 1

1 122r 2r 1 2r 2r 1

......... 2 21 1 1 1 1 121 5 5 13 2n 2n 1 2n 2n 1

2 2

2 2 2

2 2n 2n1 4n 4n2 12n 2n 1 2n 2n 1 2n 2n 1

lim lim2

n n n 24n 4nS

2n 2n 1

limn

2

44 4n 22 1 22n n

33. bUsing solution of above question

2

n 24n 4nS

2n 2n 1

. 2

16 2

4 16 4 16 1088S5452 16 2 16 1

34. aUsing solution of above question.

2

n 24n 4nS

2n 2n 1

a = 4, b = 4, c = 0, d = 2, e = 2

35. A - S , B - S, C - P, D - Q

Sol: y = exp {(tan2 x ) loge 2}= 2tan x2

since 2tan x2 satisfies the equation

x2– 17x+ 16 = 0 2tan x2 16 or 1

tan x 2 or tan x = 0

Since x (0, )2

tan x 2

(A) 22sin2x

1 cos x= 2

4 tan x 8 46 32 tan x

(B) 2sinx

sinx cos x=

2 tan x 41 tan x 3


RAMKI

(C) n

n 1(cot x)

=

cot x 11 cot x

(D) 2n

n 1n(cot x)

=

2 2

2 22

cot x cot x 491 cot x 1 cot x

36. Given 1 1 1 1

2

-----(i)

2 2 21 1 1 9

4

......(ii)

a + b + g = 2. -----(iii)

(A) from (i) 1 1 1 1

2

squaring both sides

2 2 21 1 1 12

4

or, 9 2 124 4

or, 2

(B) from equation (i) 1 1 1 1

2

or,12

or, = 2 1

2 2

(C) From (iii) 2

squaring both sidesa2 + b2 + g2 + 2 (ab + bg + ag) = 4or, a2 + b2 + g2 + 2 × –1 = 4or a2 + b2 + g2 = 6

(D) as a3 + b3 + g3 – 3abg = (a + b + g) [a2 + b2 + g2 – (ab + bg + ga)]or, a3 + b3 + g3 – 3(–2) = 14or, a3 + b3 + g3 = 8

37. Given b – a = c – b and (c – b)2 = a (b – a)

or (b – a)2 = a (b – a)

Cancel b – a 0

b – a = a or b = 2a

and c = 2b – a, by (i)

or c = 4a – a = 3a \ a, 2a, 3a or 1 : 2 : 3


RAMKI

38. Sum of first four 4a + 6d = 56

Sum of last four 4l – 6 d = 112

Putting a = 11, we get d = 2,

l 31 a n 1 d

n 1 10 or n = 111

39. 2b = a + c 2 3t t + t + 6

or t3 + t2 + t + 6 = 0

or (t + 2) (t2 – t + 3) = 0

t = –2.

The other factor does not give real values of t.

Hence for t = – 2 the given number are –10, –2, 6 which is an A.P. for d = 8.

The next two number are 14, 22.

40. 1 2

11 2

a a1Tda a

n 1 1 n1S T a ad

The terms will cancel diagonally

1 nn

1 n

a a1Sd a a

Now put n 1a a n 1 d

n1 n 1

n 1 d1 n 1d a a a a

41. . In an A.P. sum of terms equidistant from begining and end is constant and equal to a + l = a1

+ a2n. futher same as question (31)

42. Three numbers in G.P. are a ,a,arr

then a ,2a,arr

are in A.P. as given

\ 12 2a a rr


RAMKI

or, r2 – 4r + 1 = 0 or r = 2 + 3

or, r = 2 + 3 as r > 1 for an increasing G.P..

43. p = Infinite G.P.

where a = 1, r = – tan2 x

22

a 1p cos x1 r 1 tan x

, 2

2

1q sin y1 cot y

2 2

1S1 tan x cot y

2 2

2 2

11 cos x 1 sin y1cos x sin y

pq 1Sp q 1 1 1 1

p q pq

44.A B A B 22AB 2AB H

2 2nE 1 1 1 ... 1H H

45. a1, H1, H2,H3, Hm, Hn, b are in H.P.

1 n

1 1 1 1d & ndH a H a

1 1& n 1 db a

on putting values, we get value of 1 n

1 n

H a H bH a H b

46. Let T1 & T2n+1 are A & B

A B 2ABa , b AB, c2 A B

2b ac


RAMKI

47. Let tr denote the rth term of the series, then

3 3 3

r1 2 ........ rt

1 3 ........ 2r 1

222

2

1 r r 1 14 r 14r

n n n 1

2 2r

r 1 r 1 r 1

1 1t r 1 r 14 4

n 1 n 2 2n 31 14 6

21 2n 9n 13n 6 624

2n 2n 9n 1324

48. Let 1r 1t r log

r

r log r 1 logr

r 1 log r 1 r logr log r 1

n n

rr 1 r 1

t r 1 log r 1 r logr log r 1

n 1 log n 1 1log1 log n 1 !

n 1 nn 1 n 1log log

n 1 ! n!

49. We can write S as

S = (1 – 2) (1 + 2) + (3 – 4) (3 + 4) + ............ + (2001 – 2002) (2001 + 2002) + 20032

= [1 + 2 + 3 + 4 + ........... + 2002] + 20032

= 21 2002 2003 2003 20070062


RAMKI

50. Since

mm m m1 2 n 1 2 na a ..... a a a ..... a

n n

If m > 1 & not all a1, a2, a3 .......... am equal

mmm m m1 3 5 ..... 2n 1 1 3 5 ..... 2n 1n n =nm+1

51. Since AM GM

1sin x 1 sin x 1

sin x 1 sin x 1 25 5 5 .52

sin x 1 sin x 1 25 55

52. Let the four numbers be a, a + d, a + 2d, (a + d).r2.

where d is the common difference of A.P. and r is common ratio of the G.P.

a = 6 and r = 12

is given

a + d, a + 2d, (a + d) r2 are in G.P.

(a + 2d)2 = (a + d)2 r2 2 2 16 2d 6 d .4

2 24 6 2d 6 d 16 2d 6 d .2

d = –2 The 4 numbers are 6, 4, 2, 1.

53. As a912, a951 and a480 are divisible by 3, none of them is prime.

For a91, we have

91

9191 times

1 1a 99..............9 10 19 9

137 71377

10 11 10 110 19 10 110 1

12 117 7 7 6 510 10 ........ 10 1 10 10 ........ 10 1

91a is not prime.


RAMKI

54. As all xi ‘s are positive integers and 1 2 3 4x .x .x .x 64

1 2 3 4x ,x ,x ,x must be power of 2.

Then, amongst all possible values of 1 2 3 4x .x .x .x 64 and in that case x5 = 16.

55. It can be shown by grouping that n a n n2

For n = 200 we get a (200) > 100

and for n = 100, a (100) < 100

choices (a) and (d) are correct

56. Last term of nth row

= 1 + 2 + 3 + ............ + n =1/2 n (n + 1)

As terms in the nth row forms an A.P. with common difference 1,

first term = last term – (n – 1) (i)

1n n 1 n 12

21 n n 22

Sum 2 21 1 1n n n 2 n n2 2 2

21n n 12

57. Ans: (A)

Sol.The Assertion A can be proved by taking the intersection of the inequaliaties.a > 0, ar > 0, ar2 > 0, a + ar > ar2, ar + ar2 > a.ar2 + a > arThe inequalities follow from Reason

58.Ans: (A)Sol:” Statement -1

As A. M. G.M.

a b a b2b c b c

Or, a b a2b c c .....(i)


RAMKI

also c d c2d e e .....(ii)

Inequality (i) ́ (ii)

a b c d a c4b c d e c e

or.

a b c d a4b c d e e

statement (1) is correctStatement (2)As A. M. G.M.

5b c d e a b c d e a5a b c d e a b c d e

or, b c d e a 5a b c d e

So, statement (2) is correct and is correct explanation for statement (1)59.Ans: (B)Sol: Statement (1)

as a, b, c, d, are in HP‘b’ is the single H.M. between a and c

also A.M. between a and c is a c2

as, A.M. > H.M.

a c b2

a + c > 2b ........(i) ‘c’ is the single H.M. between b and d

A.M. between b and d is b d2

as , A.M. > H .M.

b d2 > c

b + d > 2c ...........(ii)inequality (i) + (ii)a +c + b+ d > 2b +2ca+ d > b +c so statment (1) is correct

Statement (2) as a, b, c, d, are in H.P. , , ,1 1 1 1a b c d will be in A.P..

1 1 1 1b a d c


RAMKI

or, 1 1 1 1a d b c

So, statement (2) is correct and is not correct explanation for statement (1)

60. ..........2 2 2 21 1 1 1S1 2 3 4

.........2 2 2 21 1 1 11 3 5 7

...........2 2 21 1 1

2 4 6

(rearrangement)

........ .........2 2 2 2 2 21 1 1 1 1 1 1S

41 3 5 1 2 3

...........2 2 21 1 1 SS

41 3 5

...........2 2 21 1 1 3 S

41 3 5

61. The method is same as in above question.

62. .........1 1 1 1101 102 103 200

......... ...........1 1 1 1 1 11 12 3 200 2 3 100

......... .......... .......1 1 1 1 1 1 1 1 1 11 1 13 5 199 2 2 3 100 2 3 100

............1 1 1 1 112 3 4 199 200

63. A-Q, B-R, C-P, D-SSol: We have by third and fourth conditions

2a = 2 + bc2 = 18bAlso a + b + c = 25On solving the system, we will get a = 5, b = 8, c = 12

64. A-R, B-P, C-Q, D-SSol: 12 – 32 + 52 – 72 + ..... – (4n – 1)2

= [12 + 52 + ....... + (4n – 3)2] – [32 + 72 + ........+ (4n – 1)2]

2 2n n 116 8n 8n 8n 8n 8n

2

B-P22 – 42 + 62 – 82 + ............ – (4n)2


RAMKI

= 22 [1 – 22+ 32 – 42 + .......... – (2n)2]= 22 {[1 + 22 + + 32 + 42 + .......... + (2n)2] – 2 [22 + 42 + 62 + ....... + (2n)2]}

= 2n 2n 1 4n 1 8n n 1 2n 1

6 6

= –2n2 – n

INTEGER ANSWER TYPE QUESTIONS1. An infinite G.P. is selected from

1 1 11, , ,2 4 8

to converge to 1/7. If 1/2a is the first term of such a G.P., find a.

2. Find the smallest natural number m > 90 for which

mtimes

n 111......1

is not a prime number. Hence find the value of m-87.

3. Suppose a, x, y, z and b are in A.P. when x + y + z = 15, and a, a, b, g, b are in H.P.. when 1/a + 1/b+ 1/g = 5/3. Find a if a > b.

4. Find

12 4

k 1

8 2ktan2 k k

5. Let a1, a2,.......,an be an A.P. with common difference p/6 and assume

sec a1 sec a2 + sec a2 sec a3 +.... + sec an–1 sec an.

= k (tan an – tan a1 )

find the value of k.

6. If the lengths of the sides of a right triangle ABC right angled at C are in A.P., find5 (sin A + sin B).

7. First term of an A.P. of non-constant terms is 3 and its second, tenth and thirty-fourth terms form aG.P., find the common difference.

8. A ball is dropped from a height of 900 cm. Each time it rebounds, it rises to 2/3 of the height it hasfallen through. Find the two times of total distance travelled by the ball before it comes to rest in decameters.

9. Find the largest positive term of the A.P. whose first two terms are 2/5 and 12/23.

10. If logx y, logz x, logy z are in G.P. , xyz = 64 and x3, y3, z3 are in A.P., find x + y + z.


RAMKI

Hints and Solutions

1. Let common ratio is12b

and

112

11 712

a

b

aSr

3&b a b

3&b a b

hence a = 3

2. for m = 91, n = (1111....91 times) is divisible by 3 since sum of digits is 91 which is divisible by 3 hencen is not a prime number.

Hence m - 87 = 4

3. Let 2 , , , , 2 a A d x A d y A z A d b A d

givin 15 5 x y z A

a 5 2d,b 5 2d

also 1 1 1 1 1, , , ,a b in A.P

1 1 1 3 1 1 52 3a b

or 3 1 1 52 5 2d 5 2d 3

d 2

take d = -2 since a > b

hence a = 9

4. 1

2 21

2tan1 1 1

n

K

kk k k k

2 2

12 2

1

1 1tan

1 1 1

n

K

k k k k

k k k k


RAMKI

1 2 1 2

1tan 1 tan 1

n

Kk k k k

1 2 1tan 1 tan 1 n n

when n

Then summation 1

2 41

2tan2 2 4 4

k

kk k

Hence 8

4

is 2

5.1 2 2 3 1

1 1 1........cos cos cos cos cos cos

n na a a a a a

2 1 3 2 1

1 2 2 3 1

sin sin sin1 .....sin cos cos cos cos cos cos

n n

n n

a a a a a ad a a a a a a

2 1 3 2 n n 11 tan a tan a tan a tan a tan a tan a

sin d

11 tan tan

sin na a

d

1 2sin

Kd

6. here 0 0C 90 ,A B 90

2 2 2c a b & 2b a c

since 2 2 2c 2b a & c a b

2 2 22b a a b

or b 4a 3

sinB 4 sinB sinA 7orsinA 3 sinB sinA 1


RAMKI

or B A 1cot

2 7

A B 7cos

2 5 2

Also 5(sinA+SinB) = A B5 2 cos

2

=7

7. 2 10 34T 3 d,T 3 9d,T 3 33d

since 2 10 34T .T ,T are in G.P2

10 2 34T T T

23 9d 3 d 3 33d

d 0,1

hence d = 1

8. According to question. total distance 2 32 2 2h 2 h 2 h 2 h ...

3 3 3

up to infinite

= 2h 2 h 33

= 5h 4500cm

10 9000 9h cm deca meters

9. Let last term is nT

n12 14T n 123 115

= 74 14n

115

hence for n = 5 last positive term obtained.

10. According to question,

3 3yz

x z

log zlog x log x log zlog y log x

x z


RAMKI

Since 3 3 3 3 32y x z x y or x y

Pivln xyz 64 & x y z

x y z 4

&x y z 12

SUBJECTIVE QUESTION

1. (i) The value of x + y +z is 15. If a, x, y, z, b are in AP while the value of 1 1 1x y z .is

53

If a,x, y, z, b

are in HP then find a and b.

(ii) If x, y, z are in H.P. then show that

log (x +z) + log (x+ z -2y) = 2 log (x - z)

2. If the mth, nrh and pth terms of an A.P. and G.P. are equal and are x. y. z then prove that

xy-z yz-xzx-y =1.

3. The intreior angles of a polygon are in A.P. the smallest angle is 1200 and the common difference is 50,Find the number of sides of the polygon.

4. Let the angles of a of triangle ABC be in A.P. and let b : c = 3 : 2 .Find the anlge A.A.

5. If n is natural number such that n = 31 2 k1 2 3 kp ,p .p .....p and p1, p2, ....pk are distinct primes, then show

that log n k log2

6. Let p be the first of the n arithmetic means between two numbers and q the first of n harmonic means

between the same numbers. Show that q does not lie between p and 2n 1 p

n 1

7. If S1, S2, S3, ......sn are the sums of infinite geometric series whose first terms are 1,2,3,......., n and

whose common ratios are 1 1 1, , ,2 3 4

..........1

n 1 respectively, then find the values of

8. If S1, S2, S3, ......sn are the sums of infinite geometric series whose first terms are 1,2,3,......., n and

whose common ratios are 1 1 1, , ,2 3 4

..........1

n 1 respectively, then find the values of

2 2 3 21 2 3 2n 1S S S ..........,s

9. The fourth power of the common difference of an arithmetic progression with integer entries in added tothe product of any four consecutive terms of it, prove that the resulting sum is square of an integer.

10. Let a1 , a2, ... be positive real number in G.P. for each n, let An, Gn, Hn be respectively, the arithmeticmean, geometric mean , harmonic mean of a1 , a2 .....an. Find an expression for the geometric mean ofG1, g2, ....gn in terms of A1, A2, ......An , H1, H2, ......Hn.


RAMKI

11. Let a and b are positive real numbers. If a, A1 .A2 , b are in arithmetic progression a, G1, G2, b are in

G.P. and a, H1, H2 ,b are in H.P. show that 1 2 1 2

1 2 1 2

2a b a 2bG G A AH ,H H H 9ab

12. If a, b and c are in arithmetic progression and a2, b2 and c2 are in Harmonic progression, then provethat either a = b = c or a, b and -c/2 are in Geometric Progreeion.

13. If a, b, c are positive numbers, then prove that (1+a)7(1+b)7 (1+c)2> 77 a4b4c4

14. If total number of runs scored in n matches is n 1

4

(2n+1-n-2) where n > 1, and the runs scored in

the kth match are given by k 2n+1-k, where 1 k n. find n

15. If 2 3 n

n 1n

3 3 3 3a ...( 1)4 4 4 4

and bn = 1 -an, then find the least natural number 0n such

that bn > an 0n n .

SUBJECTIVE QUESTIONS

Hints and Solutions1. a + b = (a +x +y +z +b) - (x + y +z)

a + b = 5 a b 15 a b 102

..........(1)

a, x, y, z, b are in HP 1 1 1 1 1, , , ,a x y z b are in APAP

1 1 1 3 1 1 5x y z 2 a b 3

a b 10ab 9

or ab = 9

2. Solving for a and b using (a-b)2 = (a+b)2 - 4ab from (1) & (2) a = 9, b = 1 or a = 1, b = 9.

Then x = a + (m -1 ) d and x = b rm-1

y = a+(n-1)d and y = b rn-1

z= a + ( p-1) d and z = b rp-1

x - y = (m-n) d, y -z = (n-p) d, z -x = ( p-m) d

Now n p d p m d m n dy z z x x y m 1 n 1 p 1x y z br br br

= b[n-p+p-m+m-n]d r[(m-1)(n-p)+(n-1) (p-m)+(m+n)]d

= b0.d r 0.d = 1


RAMKI

3. Let the number of sides of the polygon be n Then the sum of all the interior angles = (n´ 180 -360) sumof the interior angles .

= 120 + 125 + 130 .... to n terms

= n 240 n 1 52

0 0n 240 n 1 5 n 180 3602

n2 - 25n + 144 = 0

(n-9) ( n-16) = 0

n = 9 or 16

But when n = 16, the greatest interior angle is 1200 +(16-1)5 =1950 which is not possible, for interiorangle is < 180.Hence the number of sides = 9.

4. A + B + C - 1800 , 2B = A + C ( A, B, C are in A.P. )

3B = 1800

B = 600

given b c k3 2 b k 3

c k 2

By the sine rule, b c

sinB sinC

0k 3 k 2 1sinC

sinCsin60 2

C = 450

C = 1350 in impossible

A = 1800 - (B +c)

= 1800 - (600+ 450) = 750

5. Given that n = 31 2 k.1 2 3 kp .p p .......p ..........(1)

where n N and 1 2 3 kp ,p ,p .....p are distinct prime numbers

Taking log on both sides of eq. (1), we get

log n =a1 log p1 + a2 log p2 + ....+ ak log pk .........(2)

p1 2

loge ip loge 2 ......(3)


RAMKI

i 1,........k

Using (2) and (3) we get

log n a1 log 2+ a2 log 2+ a3 log 2 +......+ak log 2

log n (a1 + a2 +....+ak) log 2

log n k log 2.

6. Suppose the given two numbers to be A and B. Then the n arithmetic means are

a1, a2, a3, a4, a5,....................an

So p = A + d = A + B An 1

p = nA Bn 1

For Harmonic progression, Suppose x1, x2, ......xn to be Harmonic means then Harmonic progressionfor N +2 term is given by

A, x1,x2, x3,.....xn, B

Or q= Ist Harmonic mean = n 1 AB

nB A

Now we have to prove that q does not lies between p and 2n 1 p

n 1

So the prove the given, we have to show that q is less than p. For this

Let

2nA B nB Ap

q n 1 AB

Then

22 2 2

2

n A B AB n 1 n 1 ABp 1q n 1 AB

7. The sum of the series a + ar+ ar2 + ...........= a

1 r

from this the sum of given series = Sk =

k k 111k 1

’

Now the sum of series, here

2 2 2 2 2 21 2 3S S S ........... 2 3 .........(2n)


RAMKI

= 2 2 2 2 21 2 3 .............(2n) 1

2n 2n 1 2 2n 11

6

n 2n 1 4n 1 33

8. Since, x1, x2. x3 are the roots of x3 - x2 + bx +g = 0

we have 1x = a -d + a+a +d =1 ......(1)

1 2x x = (a -d) a+a(a+d)

+(a-d) (a +d) = b .......(2)

a b g = (a-d) a( a+d) = - g ......(3)

From (1), we get 3a = 1 a = 1/3

From (2) , we get, 3a2 -d2 = b

3(1/3)2-d2 = b 1/3 -b =d2

1 03 2d 0

13

1( , ]3

From (3), a (a2 - d2) = - g

21 1 d

3 9

21 1 d27 3

21 1 d27 3

1 0

27

127

1 ,

27

Hence b 1( , )3

and 1[ , ]

27

9. Let four consecutive terms of the A.P. be a-3d, a-d, a+d and a+3d, Common difference in 2d.

Given a -3d, a-d, a+d and a+3d are integer s . Therefore, 2d is also an integer.

Now

E= (a - 3d) (a -d) (a +d) (a +3d) +(2d)4


RAMKI

= (a2–9d2)(a2- d2) + 16d4

=a4 - 10d2 a2 + 9d4 +16d4

E = (a2 -5d2)2 is an integer

(As a-3d, a+ 3d and 2d are integers a2 -5d2 is also an integer)

Thus, E is the square of an integer

10. Let a be the first term and r be the common ration of the G.P. a1, a2, a3,....... then ak = ark-1 for k =1,2,3....As a1, a2, .... are positive real numbers

a>0 , and r > 0

I case : when r 1

we have An =1n

(a1 + a2 + .......+an)

n 11 2

1(a a ...... ar )n

n na 1 r a 1 r1n 1 r n(1 r)

Gn = (a1, a2 ...an)1n

and n 1n 1 2 n

1 1 1 1 1 1 1 1 1... ...H n a a a n a ar ar

n 1

n nn(1 r)arH

1 r

Thus An Hn = n n 1

na(1 r ) n(1 r)ar.n(1 r) 1 r

= a2 r2(n-1) = G 2n

Next, let G be geometric mean of G, G2, ...Gn then

G= (G1 G2 ...Gn) 1/n

2n 21 2 nG (G G .....G )

= (A1A2.....An) (H1H2 .....Hn)(A1H1)(A2H2)....(AnHn) 1/ 2n

1 2 n 1 2 nG (A ,A ........A H H ........H )

II case : When r = 1


RAMKI

In this case An= 1 (a a .... a) an

Similarly Gn = a and Hn = a

Also An Hn = a2 =Gn2

In this case too,

G= (A1, A2...An H1 H2 ....Hn)1/2n

11. Since a , A1, A2,b are in A.P. A1 +A2 = a + b ........(1)

a, G1, G2 b are in G.P.

1 2G G ab

a, H1, H2 b are in HP 1 23ab 3abH ,H

2b a b 2a

1 2 1 2

1 2 1 2

H H A Aa bH H ab G G

from (1) and (2)

1 2 1 2

1 2 1 2

G G A AH H H H

........(3)

we also know a, H1 , H2, b, are in H.P.

1

1 1 1 1 1H a 3 b a

we get 13abH

2b a

and similarly for 2

3abH2a b

sub stituting (3), we get desired result

1 2

1 2

2b a 2a bA AH H 9ab

Hence proved.

12. Given that 2b = a+ c ......(1)

a2 , b2 c2 are in H.P.

and 2 2

22 2

2a cba c

........(2)

From (2) 2 2

22

2a cb4b 2ac

using (1)


RAMKI

2 2(ac b )(ac 2b ) 0

2b ac or 2b2 = -ac

Case I : b2 = ac

2a c ac2

using(1)

a c

a = b= c, as a, b,, c, are in A.P..

Case II : 2b2 = -ac

a, b, -c /2 are in G.P..

13. A.M. G.M.

1

4 4 7a b c ab bc ca abc a b c7

a +b+c +ab+bc+ca+abc 7(a4b4c4)1/7

1 + a + b + c + ab + bc + ca + abc 7(a4b4c4)1/7

(1+a) (1+b) (1+c) 7(a4b4c4)1/7

)(1 +a)7(1+b)7(1+c)7 77(a4b4c4)

14.n n

n 1 k n 1 kn

k 1 k 1S k.2 2 k.2

nn 1

1 1112 22 1 11 1

2 2

n 2 n 2n n 1

1 n2 1 2 4 2n2 2

so, Sn= n 1n 1(2 2 n)4

(as given)

So, n 1 n 1n 1 2 2 n 2 2 n 2

4

n +1 = 8 , n =7


RAMKI

15. an =

2 3 nn 13 3 3 3.... 1

4 4 4 4

n

n

3 314 4 3 313 7 41

4

bn = 1 -an and bn

> an 0n n

n6 31 17 4

n

n 1 2n 13 1 ( 3) 24 6

for n to be even, inequality always holds. For n to be odd, it holds for n 7.

The least natural number for which it holds is 6

( it holds for every even natural numbers)

ADDITIONAL QUESTIONS


1. The sum of the two numbers is 126

. An even numbers of arithmetic means are inserted between themand their sum exceeds their number by 1. Then the number of means inserted is

(A) 6 (B) 8

(C) 12 (D) 15

2. If a and b are the roots of the equation x2 – ax + b = 0 and vn = an + bn, alsovn+1 = avn – bvn–1 then the value of a5 + b5 is .

(A) a5 +5a3 b + 5 ab2 (B) b5 + 5b3a +5a2b

(C) a5 - 5a3b + 5ab2 (D) a5 + b5 +5ab

3. 111 ....... 1 (91 times) is a

(A) Prime number (B) Conposite number

(C) Not a integer (D) None of these

4. The sum of16 terms of the series

18

8 16 128.........5 65 2 1

is


RAMKI

(A) 540

1088(B)

1088545

(C) 1001500

(D) 1013545

5. If 1 1 1f r 1 ......... and f 0 02 3 r

, then value of n

r 12r 1 f r

(A) n2 f(n) (B) (n + 1)2 f (n+1) 2n 3n 2

2

(C) (n +1)2 f(n) 2n n 1

2

(D) None of these

6. Two consecutive numbers from 1, 2, 3, ........ n are removed Arithmetic mean of the remaining numbers

1054

then n and those removed numbers are.

(A) 7, 8 40 B) 50 , 7,8

(C) 50, 15 ,6 (D) 40,4, 6

7. A sequence a1,a2,a3, ........... an of real numbers such that

1 2 1 3 2 n n 1a 0, a a 1, a a 1,......., a a 1

then the arithmetic mean 1 2 na a ........... a

n

of these numbers can not be less than

(A) -1 (B) 1

2

(C) -2 (D) -3

8. Solution of the system of equation

2x4 = y4 + 24, xyz = 8

knowing that the logarithms logy x, logz y, logx z form a geometric progression is

(A) x = y = =z = 2 (B) x = y = 2 , z = 3

(C) x = 2, y = z = 3 (D) x = y = z= 3

9. Given that a1, a2, .......... an form an arithmetic progression with common difference 2 &a1 = 1 then thefollowing sum

10

i i 1 i 2

i 1 i i

aa aSa a 2 is

(A) 920 (B) 865

(C) 720 (D) 635


RAMKI

10. If a, b, c be distinct positive in G.P. and logc a, logb c, loga b be in A.P., then common difference of thisA.P. is .

(A) 3/2 (B) 1/2

(C) 2 (D) 5/2

11. A three digit number whose consecutive numbers form a G.P. If we subtract 792 from this number, weget a number consisting of the same digits written in the reverse order. Now if we increase the seconddigit of the required number by 2, the resulting number will form an A.P. Then the number is

(A) 139 (B) 927

(C)931 (D) 763

12. A man invests Rs. P at the end of the first year, Rs. 2P at the end of the second year,Rs. 3P at the end of the third year, and so. on, upto the end of the nth year. If the rate of interest in Rs.r per and the interest is compounded annually, then the amount the man will receive at the end of the(n + 1) year is

(A)

2

2

p r 1r

(B)

n2

2

p r 1) r 1 1 np r 1rr

(C)

n2

2

p r 1) r 1 1

r(D)

np r 1r

13. The nth term of a series is given by

5 3

n 4 2

n ntn n 1

and if sum of its n terms can be expressed as

2n n 2

n

1S a ab b

, where an and bn are the nth terms of some arithmetic progressions and a, b are

some constant, then n

n

ba equal to.

(A) n 2 (B) n2

(C) 12

(D) 2

More than One Correct Answer Type Questions14. The numbers 1, 5, 25 can be three terms (not necessarily consecutive) of

(A) at least one A.P. (B) at least one G.P.

(C) infinite number of A.P’s (D) infinite number of G.P.’s


RAMKI

15 If the a, b, g are the roots of an equation x3 + bx2 + 3x – 1 = 0 ( , , , are in H.P.) then

(A) one of the roots must be 1

(B) one root is smaller than 1, other is greater than 1

(C) b 3,

(D) all the roots must be equal

16. Three positive real numbers a, b, c are such that a2, b2, c2 are terms of an arithmetic progression. Then

(A) a, b, c are terms of a geometric progression

(B) (b + c), (c + a), (a + b) are terms of a geometric progression

(C) (b + c), (c + a), (a + b) are terms of a harmonic progression

(D) the condition of a, b, c being positive numbers may be withdrawn

17. The natural numbers are divided into groups 1; 2, 3, 4; 5, 6, 7, 8, 9; 10, 11, 12, 13, 14, 15, 16 and so on,then

(A) 1st number of 10th group is 80

(B) 1st number of the 10th group is 82

(C) 10th group will contain 19 terms

(D) Sum of the elements in 10th group will be 1729

18. Let n n2 2 2 2

1 1 1 1 1S ........... and T 2n,1 2 3 n

then

(A) S2 < T2 (B) If Sk < Tk then SK + 1 < Tk+1

(C) Sn < Tn for all n 2 (D) Sn > Tn for all n 2007

19. Let n 11 1 1n 1 . ......... 12 3 n

(A) 1 1 1........... 2nn 1 n 2 2n

(B) 2n 1 n

(C) 2n 0.5 n (D) 0.5 n 1 n

20. Let a sequence {an} be defined by

n1 1 1 1a .............

n 1 n 2 n 3 3n

(A) 27a

12 (B) 2

19a20

(C) n 1 n9n 5a a

3n 1 3n 2 3n 3

(D) n 1 n2a a

3 n 1


RAMKI

21. If |x| < 1, then 1 + 23x + 33x2 + 43x3 + ...... + n3xn–1 + ...... must be

(A)

2

3

1 4x x1 x

(B)

2

4

1 4x x1 x

(C) rational function of x (D) a polynomial function of x

22 A sequence {an} is defined as a0 = 9 and an+1 4 3n n3a 4a . Then

(A) a2 + 1 is divisible by 104 (B) a3 s divisible by 38

(C) an + 1 is divisible by n210 (D) an is divisible by 32n

23. The sequence {xn} satisfies x1 = 0 and 2n 1 n nx 5x 24x 1 . Then

(A) xn is rational for all n (B) xn is rational for all n 24

(C) xn+1 = 10xn – xn–1 (D) xn is positive integer for all n


24. STATEMENT-1 : The maximum number of acute angles in a convex polygon of n sides is 3

STATEMENT-2 : The sum of internal angles of any convex polygon is (n – 2) 1800

25. Suppose four distinct positive numbers a1, a2, a3, a4 are in G.P.

Let b1 = a1, b2 = b1 + a2, b3 = b2 + a3 and b4 = b3 + a4.

STATEMENT-1 : The numbers b1, b2, b3, b4 are neither in A.P. nor in G.P.

STATEMENT-2 : The numbers b1, b2, b3, b4 are in H.P.

26. STATEMENT-1 :

2 2 2 n n 11 2 n............1 3 3 5 2n 1 2n 1 2 2n 1

because

STATEMENT-2 : 1 1 1 1.........

1 3 3 5 2n 1 2n 1 2n 1

27. STATEMENT-1 : k k kkfor k 0, let S n 1 2 .......... n ,

then 24

1S n n n 1 2n 1 3n n 130

STATEMENT-2 : k 1k k 1 1 0

k 1 k 1 k 1 k 1S n S n ......... S n S n n 1 1

1 2 k k 1


RAMKI

28. Let a0, a1, a2, a3, ........... be an arithmetic progression

STATEMENT-1 : sin a2 + sin a4 + ............ + sin a2n 1 2n 1

2 1

cosa cos a2sin a a

STATEMENT-2 : cos a2 + cos a4 + ........... + cos a2n = 2n 1 1

2 1

cos a cosa2cos a a

Comprehension Type QuestionThis section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered.Each question has 4 choices (A), (B), (C) aP29–31 : Paragraph for Question Nos. 29 to 31

The numbers, 1, 3, 6, 10, 15, 21, 28, ............ are called triangular numbers. Let tn denotes the nthtriangular number then it can be observed that t1=1, t2 = 3, tn = tn–1+ n. Answer the following questions.

29. t100 must be equal to

(A) 5050 (B) 5151

(C) 5252 (D) None of these

30. If m is the nth triangular number then

(A) 1 8m 1n

2

(B) 1 8m 1n

2

(C) 1 4m 1n

2

(D) None of these

31. The number of positive integers lying between t50 and t51 must be

(A) 50 (B) 51

(C) 52 (D) None of theseP32–34 : Paragraph for Question Nos. 32 to 34

Let vr denote the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and thecommon difference is (2r – 1). Let Tr = Vr+1 – Vr–2.

r r 1 r rr

1 1Q T T ,WT 4 r 1

and

rQrX 3 for r 1,2,...

32. The sum 1 2 nV V ... V is:

(A) 21 n n 1 3n n 112

(B) 21 n 1 3n n 212

(C) 21n 2n n 12

(D) 31 2n 2n 33


RAMKI

33. Tr is always :

(A) an odd number (B) an even number

(C) a prime number (D) a composite number

34. Which one of the following is a correct statement ?

(A) Q1 , Q2, Q3, ..... are in A.P. with common difference 5.

(B) Q1 , Q2, Q3, ..... are in A.P. with common difference 6.

(C) Q1 , Q2, Q3, ..... are in A.P. with common difference 11.

(D) Q1 = Q2= Q3 = ....P35–37 : Paragraph for Question Nos. 35 to 37

Let ABCD is a unit square and 0 < a < 1. Each side of the square is divided in the ratio a : 1 – a, as shownin figure. These points are connected to obtain another square. The sides of new square are divided in theratio a : 1 – a and points are joined to obtain another square. The process is continued indefinitely. Let an

denote the length of side and An the area of the nth square

35. The value of a for which nn 1

8A3

is

(A) 1/3, 2/3 (B) 1/4, 3/4

(C) 1/5, 4/5 (D) 1/2

36. The value of a for which side of nth square equals the diagonals of (n + 1)th square is

(A) 1/3 (B) 1/4

(C) 1/2 (D) 1/ 2

37. If a = 1/4 and Pn denotes the perimeter of the nth square then nn 1

P

equals

(A) 8/3 (B) 32/3

(C) 16/3 (D) none of these


RAMKI


This section contains 2questions. Each question contains statements given in two column which have to bematched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II.

The answers to these questions have to be appropriately bubbles as illustrated in the following example.If thecorrect matches are A–p, A–s, B–q, B–r, C–p, C–q and D–s, then the correctly bubbled 4 × 4 matrix should beas follows :

______________________________________________________________________________

38. Match the sequence a1, a2, a3,...... whose nth term is given on the left with properties of the squenceon the right.

Column I Column II

(A) an = / 2

0

1 cos2nx dx1 cos2x

(p) a1, a2, a3.... are in A. P.

(B) In = / 4 n

0tan x dx

and nn 1 n 3

1aI I

(q) a1, a2, a3.... in G. P

(C) an = 2/ 4

0

sin nx dxsin x

(r) a1, a2, a3.... are in H. P

n 0

and n n n 1a I I n 1 .

(D) 2/2

n 20

sin nxb dxsin x

(s) a1, a2, a3.... is a constant sequence.

n n n 1a b b n 1 (t) a1, a2, a3.... is A.G.P.

39. Let a1, a2, a3, ...... be a geometric progression such that

10 m 10 n1 1log a and log an m

for two fixed positive integer m and n, with m < n, thenColumn I Column II(A) a2m+n (p) 101/m–1/n

s(B) amn (q) 10(C) am+n (r) 102/n+1/m

(D) an–m (s) 101/n+1/m


RAMKI

ADDITIONAL QUESTIONS

(ANSWER KEY)

1-C 2-C 3-B 4-B 5-B 6-B

7-B 8-A 9-B 10-A 11-C 12-B

13-D 14-A,B,C,D 15-A,D 16-C,D 17-B,C,D 18-B,C

19-A,B,C 20-B,C 21-B,C 22-A,C 23-A,C,D 24-B

25-C 26-C 27-D 28-C 29-A 30-B

31-A 32-A 33-D 34-B 35-B 36-C

37-D

38- A P BP CR DP,Q,R,S

39. A R BQ C S DP

Hints And SolutionsAdditional Questions

1. Let the two numbers a and b

given 13a b6

and A.M.’ s are A1, A2, ......... A2n inserted between a and b.

Here a, A1, A2, ..................., A2n, b are in A.P. then given condition

A1 + A2 + ......... + A2n = 2n + 1

or (a + A1 + A2 + ......... + A2n + b) – (a + b) = 2n + 1

or 2n 2

a b a b 2n 12

or n (a + b) = 2n + 1

or 13n = 12n + 6

or n = 6

Hence number of means are 12

2. a + b = a, ab = b and vn = an + bn

vn+1 = an+1 + bn+1 \ v2 = a2 – 2b

= (a + b) (an + bn) – ab (an–1 + bn–1)


RAMKI

vn+1 = avn – bvn–1 ....... (i)

v5 = a5 + b5 = av4 – bv3

= a [av3 – bv2] – bv3

= (a2 – b) v3 – abv2

= (a2 – b) [av2 – bv1] – abv2

= (a3 – 2ab) (a2 – 2b) – ab (a2 – b)

Hence a5 + b5 = a5 – 5a3b + 5ab2.

3. We have

1111 + .......... 1 (91 times)

= 1 + 10 + 102 + .......... + 1090

9110 110 1

791

7

10 110 1.

10 110 1

137 7

7

10 1 10 1.

10 110 1

= (1084 + 1077 + 1070 + ........ 107 + 1) (106 + 105 + 104 + 103 + 102 + 10 + 1)

= product of two integers (> 2). so is not a prime number.

4. Since n

n 4r 1

8rS4r 1

n

2 2r 1

8r2r 2r 1 2r 2r 1

2 2n

2 2r 1

2r 2r 1 2r 2r 12

2r 2r 1 2r 2r 1

n

2 2r 1

1 122r 2r 1 2r 2r 1

2 2

1 1 1 1 1 12 .......1 5 5 13 2n 2n 1 2n 2n 1


RAMKI

2

12 12n 2n 1

2 2

2 2

2 2n 2n 4n 4n2n 2n 1 2n 2n 1

2

16 2

4 16 4 16 1088S5452 16 2 16 1

5. Since n

r 12r 1 f r

n

2 2

r 1r 2r 1 r f r

n

2 2

r 1r 1 r f r

n

2 2 2 2

r 1r 1 f r r 1 f r 1 r 1 f r 1 r f r

n n

2 2 2

r 1 r 1r 1 f r f r 1 r 1 f r 1 r f r

2n n 1 n2 2 2

r 1 r 1 r 1

r 1r 1 f r 1 n 1 f n 1 r f r

r 1

n

2 2 2

r 1r 1 2 f 2 3 f 3 ........ n f n

2 2 2 2 2n 1 f n 1 1 f 1 2 f 2 3 f 3 ......... n f n

n n

2 2

r 1 r 1r 1 n 1 f n 1 1 f 1

2n n 1n n 1 f n 1 f 1

2

2 n n 3n 1 f n 1 1

2

22 n 3n 2

n 1 f n 12


RAMKI

6. Let p and (p + 1) be removed number from 1, 2, .........., n then sum of remaining numbers

n n 12p 1

2

From given condition

n n 12p 1105 2

4 n 2

22n 103n 8p 206 0

since n and p are integers so n must be even let n = 2r

we get 24r 103 1 r

p4

Since p is an integer then (1 – r) must be divisible by 4. Let r = 1 + 4t,

we get

n = 8t + 2 and p = 16t2 – 95t + 1, Now 1 p n

21 16t 95t 1 8t 2

t 6

n 50 and p 7

Hence removed no. are 7 & 8.

7. Let us add one more number an+1, to the given sequence. The number an+1 is such that

n n 1a 1 a . squaring all the numbers, we have

21a 0

2 22 1 1a a 2a 1

2 23 2 2a a 2a 1

..................................

...................................

2 2n n 1 n 1a a 2a 1

2 2n 1 n na a 2a 1

Adding, we get


RAMKI

n 1 n n2 2i i i

i 1 i 1 i 1a a 2 a n

n2

i n 1i 1

2 a n a n

n

ii 1

na2

1 2 na a ......... a 1n 2

8. y z xlog x,log y,log z are in G.P..

2z y xlog y log x log z

or 2z

z

1log ylog y

or 3z zlog y 1 log y 1

it is possible when y = z (x, y, z > 0)

from 2x4 = y4 + z4

2x4 = 2y4

x = y = z

from xyz = 8,

x3 = 8

x 2 x y z 2

9. 1 2 na ,a ,.........,a are in A.P..

i 2 i 1 i 1 ia a a a d (say)

i i 2i 1

a a a2

n n

i i 1 i 2i i 2

i 1 i 1i i 2

aa a 1S aaa a 2

n n

2 2 2 2 2i 1 1 1

i 1 i 1

1 1S a d a 2a di i 1 d2 2


RAMKI

2 2 21 1

n n 1 n n 1 2n 11 na 2a d nd d2 2 6

2 21 1

n 1 2n 5n a a d n 1 d2 6

10. Let r be the common ratio then b = ar, c = ar2 and logca, logbc, logab are in A.P.

e e e

e e e

log a log c log b, ,

log c log b log a are in A.P..

2

ee e2

e ee

log arlog a log ar, ,log ar log alog ar

are in A.P..

so e e e e e

e e e e e

log a log a 2log r log a log r, ,

log a 2log r log a log r log a

are in A.P..

Putting e

e

log rx

log a

we get

1 1 2x, ,1 x1 2x 1 x

are in A.P..

2 1 2x 11 x1 x 1 2x

3 22x 3x 3x 0

since a, b, c are distinct so r 1 , so x 0

22x 3x 3 0

1x 3 334

then 11 x 3

1 2x

, so the common differences of A.P. is 3/2.

11. Let the three digit be a, ar, ar2 then according to hypotesis

100a + 10 ar + ar2 + 792 = 100 ar2 + 10 ar + a , i.e. a (r2 – 1) = 8 and a, ar + 2, ar2 are in A.P.

then 2 (ar + 2) = a + ar2

so r = 3 & a = 1. Thus digits are 1, 3, 9 and so the required number is 931.


RAMKI

12. An amount of Rs. P will earn interest for n years. Therefore Rs. P will become

P (1 + r)n at the end of (n + 1) years. Rs. 2P will earn interest for (n – 1) years. Therefore

Rs. 2P will become 2P (1 + r)n–1 at the end of (n + 1) years and so on. Thus, the amount the man willrecieve is

P (1 + r)n + 2P (1 + r)n–1 + ......... + nP (1 + r)

Let S = P (1 + r)n + 2P (1 + r)n–1 + .......... nP (1 + r)

n 1 n 2S P 1 r 2 1 r ....... n 1 1 r n1 r

Subtracting we get

n n 1Sr P 1 r P 1 r ......... P 1 r nPr 1

nP 1 r 1 r 1nP

r

2 n

2

P r 1 r 1 nP r 1S

rr

13. Since,

5 3

n 4 2 4 2

n n nt nn n 1 n n 1

2 2

1 1n2 n n 1 2 n n 1

\ sum of n terms n

n nn 1

S t

n

n 2 2n 1

1 1 1S n2 n n 1 n n 1

2 2

1 1 1 1 1 1 1 11 2 3 ........ n 1 .....2 3 7 3 13 7 n n 1 n n 1

2

n n 1 1 112 2 n n 1

2

2

n n 1 1 1.2 2 2 2 n n 1

2

2n 1 1 1 1

8 2 2n 2n 12 2 2


RAMKI

2

2n 1 5 1

82 2 2 1 3n 222

but given 2

n n 2n

1S a ab b

On comparing we get

n nn 1 5 1 3a , a ,b n 2 ,b

8 22 2 2 2

Hence, n

n

a 1b 2

, which is constant.

14. Let 1, 5, 25 be the pth, qth, rth terms of an A.P. with common difference d, then

(q – p) d = 5 – 1 and (r – p) d = 25 – 1

q p r p k say1 6

q p k, r p 6k where k is any natural number..

Let 1, 5, 25 be the pth, qth, rth terms of a G.P., with common ratio R, then

Rq–p = 5, Rr–p = 25

r p 2q 2p r p 2q 0

There exist infinitely many triplets of natural numbers satisfying this relation.

15 Take, 1 1 1, ,

p q p p q

Then S = 1 1 1 b

p q p p q

........(i)

1 1 1 3

p p q p p q p q p q

........(ii)

and

1P 1p q p p q

.....(iii)

From (ii) and (iii) we easily get p = 1, whence from (iii) q = 0 1

from (i), b = –3


RAMKI

16. Given 2 2 2a c 2b

Consider, 2 1 1

c a a b b c

= 2 2 22ab 2ac 2bc 2b bc ab ac c ac a bc abc a a b a c

= 2 2 22b c a 0

c a a b b c

2 2 2a c 2b

2 1 1

c a a b b c

b c , c a , a b are in H.P..

Also, we can withdraw the given condition of a, b, c being positive.

17. First group is ending at 12

Second group is ending at 22

Third group is ending at 32 and so on.

9th group is ending at 92

first element of 10th group is 82

Now, 1st, second and third group are having 1, 3, 5, 7, ........... elements

(an A.P. whose nth term is 2n – 1)

10th group will contain 19 element

Now sum of elements in 10th group

= 82 + 83 + 84 + .......... upto 19 terms

19 2 82 19 1 17292

18. 2 22 2

1 1 5 3S ,T4 21 2

23S2

(a) is true

If Sk < Tk

Then 2 2 2

1 1 1 1............ 2k1 2 k


RAMKI

on adding 2

11 k on both side

2 2 2 2 2

1 1 1 1 1 1.......... 2k1 2 k 1 k k 1

Now 2

1 1 12 2k k 1k 1

will be true if

2

1 1 1k k 1k 1

or (k + 1)2 + k k (k + 1)

or k2 + 3k + 1 k2 + k which is true.

k 1 k 1S T

19.1 1 1..........

n 1 n 2 2n

1 1 1 1 11 ........... 1 .......2 3 2n 2 n

1 1 1 1 1 1 11 ....... 1 ......... 1 .........3 2n 1 2 2 n 2 n

1 1 1 1 11 ...........2 3 4 2n 1 2n

Also,

n times

1 1 12n .......... 1n n n

and

n times

1 1 1 12n .......... n2n 2n 2n 2

(d) is not correct as a (1) = 1

20. an is sum of reciprocals of natural numbers starting at n + 1 and ending at 3n

\ 21 1 1 1 19a3 4 5 6 20

\ n 11 1 1 1 1 1a ....

n 2 n 3 3n 3n 1 3n 2 3n 3


RAMKI

n 1 n1 1 1 1a a

3n 1 3n 2 3n 3 n 1

= 1 1 2

3n 1 3n 2 2n 3

1 1 23n 1 3n 2 3n 3

9n 5

3n 1 3n 2 3n 3

21. 3 3 2 3 n 1S 1 2 x 3 x ... n x ......(i)

3 2 3 3 3 n 1 3 nSx x 2 x 3 x ... (n 1) x n x ......(ii)

equation (ii) – (i) we get

n 2n3 n 2 n

2 3 4

1 x x 4x 13nx x 1n x 3n xSx 1 x 1 x 1 1 x

For |x| < 1, xn ® as n ®

Sum of the series is

2

4

x 4x 11 x

22 Consider 4 3 31 0 0a 3.a 4a 9 27 4 22599

1a 1 22600 which is divisible by 102

Suppose, ka 1 is divisible by k210

Let k2ka .10 1

k k4 32 2

k 1a 3 10 1 4 10 1 = k k3

2 2.10 1 [3 10 3 4]

Put, k 1k2 210 t 10

= k k 22 .2 2 210 10 t ......(i)

3 3 2 2k 1a 1 t 3 t 3 t 1 3 t 1 1

= 4 4 3 3 2 2 3 3 2 23 t 9 t 9 t 3 t t 3 t 3 t 1 1

2 4 2 3 2t [3 t 8 t 6 ]


RAMKI

k 1 k 1 k2 4 2 3 2 210 3 .10 8 .10 6

k 1a 1 is divisible by k 1210

Hence, by principle of induction, result holds for all n.

23. 2n 1 n n 1x 5x 24x 1, x 0

Note that nx is an increasing sequence and nx 0 for all n.

2 2 2n 1 n 1 n n nx 10x x 25x 24x 1

2 2n 1 n n 1 nx 10x x x 1 0

Replace n by n – 1

2 2n n 1 n n 1x 10x x x 1 0

Thus, for n > 2, xn+1 and xn–1 gives 2 distinct root of

2 2n nx 10x x x 1 0

Sum of the roots is n 1 n 1 nx x 10x

n 1 n n 1x 10x x

Also 1 2 3x 0 x 1 x 10

24. Ans (B)

Sol.Let m be the number of acute angles in a convex polygon of n-sidesThen sum of interanl angles of polygon

0 0 0n 2 180 m90 n m 180

. .,4 m i e m 4 Maximum possible value of m is 3 Statement (1) is trueStatement (2) is also true, but it does not provide the explanation for statement (1)

25. Ans: (C)

Sol: Let a be the first term and r be the common ratio of the G.P. As four numbers are distinct and positive a> 0, r > 0 and r 1

b1 = a, b2 = a (1 + r), b3 = a (1 + r + r2)

and b4 = a (1 + r + r2 + r3)

As 22 1 3 2b b ar ar b b

b1, b2, b3, b4 are not in A.P..


RAMKI

232

1 2

bb 1 r r1 rb 1 r b

b1, b2, b3, b4 are not in G.P..

As 2

22 1 3 2

1 1 r r 1 1b b a 1 r b ba 1 r 1 r r

b1, b2, b3, b4 cannot be in H.P.

26.

Ans: (C)

Sol: Let 2

rrt

2r 1 2r 1

2

r4r 1 14t

2r 1 2r 1

11

2r 1 2r 1

1 1 112 2r 1 2r 1

n n

rr 1 r 1

1 1 14 t n2 2r 1 2r 1

1 1n 12 2n 1

2n n 1nn2n 1 2n 1

n

rr 1

n n 1t

2 2n 1

\ Statement –1 is true

But statement-2 is false since

n n

r 1 r 1

1 1 1 1 1 1 n12r 1 2r 1 2 2r 1 2r 1 2 2n 1 2n 1

27.

Ans: (D)

Sol: We have


RAMKI

k 1 k 1 k k 1 0k 1 k 1 k 1 k 1r 1 r r r ..... r r

1 2 k k 1

n

k 1 k 1 k 1

r 1n 1 1 r 1 r

k k 1 1 0

k 1 k 1 k 1 k 1S n S n ...... S n S n

1 2 k k 1

\ Statement-2 is true

That statement-1 is false can be checked by putting n = 4 in

In fact

24

1S n n n 1 2n 1 3n 3n 130

28.

Ans (C)

Sol. 2 4 6 2nsina sina sina ......sina

1 1 1 2n 12sind sin a d sin a 3d sin a 5d ...... sin a d2sind

1 2n 1

2 1

cosa cosa2sin a a

stat. 2 similar given expansion becomes

2n 1 1

2 1

sin a sina2sin a a

Sol:29

Let S = 1 + 3 + 6 + 10 + 15 + ....... + tnS = 1 + 3 + 6 + 10 + ........ + tn–1 + tnSubtractions0 = 1 + 2 + 3 + ........ + n–tn

n

n n 1t

2

100100 101t 5050

2


RAMKI

30. By trial methodchoice aUse 6 as 3rd triangular numberPut n = 3 and m = 6

1 8 6 132

Choice (a) is not correctchoice (b)

1 8 6 132

Choice (b) is correct31.

5151 52t 1326

2

5050 51t 1275

2

Integer between 1326 and 1275 is 50


Sol. 32. rrV 2 r r 1 2r 12 3 21 2r r r

2

n n n n3 2

kk 1 k 1 k 1 k 1

1 1V k k k2 2

221 1 1n n 1 n n 1 2n 1 n n 14 12 4

1 n n 1 3n n 1 2n 1 312

21 n n 1 3n n 212

Sol. 33. Tr = Vr+1 – Vr – 2

3 23 21r 1 r [ r 1 r ]2

2 1 13r 3r 1 2r 1 22 2

23r r 1 3r 1 r 1 rT is always a composite number

Sol. 34. Qr = Tr+1 – Tr = 3 {(r + 1)2 – r2} + 2 {(r+1) – r}

= 3 (2r + 1) + 2 = 5 + 6r

\ Q1 , Q2,..... forms an A.P. with common difference 6.


RAMKI


35.Sol: n2 21 n 1a 1,a 2 2 1

n 2n 1

1 8A32 2

22 2 3/8

216 16 3 0

1/ 4, 3 / 4

36

Sol: Diagonal of (n + 1)th square = n 12a

2 2n n 1a 2a

2 2 2n na 2 2 2 1 a

1/ 2

37.

Sol: P1 = 4

n 1 / 22n nP 4a 4 1 2 2

n 1 / 2548

n

n 1

8 2 8 54 8P38 5

8 4 103

38.

Ans : A-p, B - p, C - r, D-p,q, r, s

Sol.1 (a)/ 2

1 20a dx ,a

2

Also, n 2 n n 1a a 2a

= / 2

0

2cos 2n 2 x cos 2n 4 x cos 2nxdx

1 cos2x


RAMKI

= / 2

0

2cos 2n 2 x 2cos 2n 4 xcos 2xdx

1 cos2x

= / 2

02cos 2n 2 xdx

/ 2

0

2 sin 2n 2 x 02n 2

1 2 3a ,a ,a ,... are in A.P..

(B) n 1 n 3I I

/ 4 n 1 2

0tan x sec x dx

/ 4n 2

0

1 1tan xn 2 n 2

na n 2

1 2a ,a ,.... are in A.P..

(C) 2 2

/2

n n 1 0

sin nx sin n 1 xI I dx

sinx

/ 2

0

sin 2n 1 x sin xdx

sin x

/ 2

0

1 1cos 2n 1 x2n 1 2n 1

1 2 3a ,a ,a , .... are in H.P..

(D)/ 2

n 0

1 cos2nx nb dx1 cos2x 2

na n2

\ a1, a2, a3,.... are in A.P,, G.P. and H.P.Also a1, a2, a3....... is a constant sequence.

39. Ans: A-r, B-q, C-s, D-pSol: Let tr = log10 ar, then t1, t2, t3, ......... are in A.P. Let its common difference be d. Now

m n1 1t tn m

m n 1m n d dmn mn

Now, r mr 1t t r m d

mn n

rt r / mn r / mnra 10

* * *

PS

Documents

appropriately

2n 1pn 1

andwhose common

bb aa ba

nr 11a rx

1m nt ar

b2 c2

r2