English Class X Maths Chapter01

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After learning this chapter you will be able to :

Set Theory* Solve the problems relating to properties of union and intersection of sets

* Understand the De Morgan’s Laws of sets

* Solve the problems relating to the relation between number of elements of sets

Sequence

* Understand the meaning of sequence and series

* Identify A.P, G.P and H.P.

* Identify the specified terms of A.P, G.P and H.P.

* Find the sum of terms in A.P and G.P.

* Find the relation between A.M.,G.M. and H.M.

Matrices

* Recall the matricies and the transpose of a matrix

* Understand the conditions necessary for addition, subtraction and multiplication ofmatrices.

SET THEORY

You have already learnt about the sets and their operations. Now we shall studymore about the properties of sets.

1. Properties of Union and Intersection of sets

(i) Commutative Property :Ex 1 : If A = { x : x ∈ N and 1 < x < 5} and B = {0, 2, 3}

Find a) A ∪ B b) B ∪ A and compare both the sets.

Solution : a) A = {x : x ∈ N and 1 < x < 5}

A = {2, 3, 4} and B = {0, 2, 3}

A ∪ B = {2, 3, 4} ∪ {0, 2, 3)

1 NUMBER SYSTEM

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∴ A ∪ B = {0, 2, 3, 4} ................ (i)

B ∪ A = {0, 2, 3} ∪ {2, 3, 4}

∴ B ∪ A = {0, 2, 3, 4} ............. (ii)

From (i) and (ii) A ∪ B = B ∪ A

Union of Sets is Commutative

Ex 2 : If A = {r, a, t} and B = {c, a, t}

Verify that A ∩ B = B ∩ A.

Solution : A∩ B = {r, a, t} ∩ {c, a, t}

∴ A∩ B = {a, t} ............. (i)

B∩ A = {c, a, t} ∩ {r, a, t}∴ B∩ A = {a, t} ............... (ii)From (i) and (ii) A∩ B = B ∩ A.

Intersection of sets is Commutative

(ii) Associative Property :Ex 1 : If A = {1, 2, 3, 4, 5, 6} B = {2, 3, 7, 10} and C = {3, 6, 7, 8, 9} then

find i) A ∪ (B ∪ C) ii) (A ∪ B) ∪ C and Verify that A ∪ (B ∪ C) = (A ∪ B) ∪ C

A ∪ B B ∪ A

B ∩ AA∩ B

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Solution : (i) To find A ∪ (B ∪ C)

B ∪ C = {2, 3, 7, 10} ∪ {3, 6, 7, 8, 9}

B ∪ C = {2, 3, 6, 7, 8, 9, 10}

A ∪ (B ∪ C) = {1, 2, 3, 4, 5, 6} ∪ {2, 3, 6, 7, 8, 9, 10}

∴ A ∪ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} .............. (i)

(ii) To find (A ∪ B) ∪ C

A ∪ B = {1, 2, 3, 4, 5, 6} ∪ {2, 3, 7, 10}

A ∪ B = {1, 2, 3, 4, 5, 6, 7, 10}

(A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7, 10} ∪ {3, 6, 7, 8, 9}

∴ (A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ................. (ii)

From (i) and (ii) A ∪ (B ∪ C) = (A ∪ B) ∪ C

∴ Union of sets is associative

Ex 2 : If A = {3, 4, 5, 9}, B = {4, 5, 6, 8} and C = {5, 7, 8, 9}

Is the Intersection of these sets Associative?

Solution : (i) To find A ∩ (B ∩ C)

B∩ C = {4, 5, 6, 8}∩ {5, 7, 8, 9}

B∩ C = {5, 8}

A∩ (B∩ C) = {3, 4, 5, 9}∩ {5, 8}

∴ A∩ (B∩ C) = {5} .............. (i)

(ii) To find (A∩ B)∩ C

A∩ B = {3, 4, 5, 9}∩ {4, 5, 6, 8}

A∩ B = {4, 5}

(A∩ B)∩ C = {4, 5}∩ {5, 7, 8, 9}

∴ (A ∩ B)∩ C = {5}............... (ii)

From (i) and (ii) A∩ (B∩ C) = (A∩ B)∩ C

∴ Intersection of sets is associative

(iii) Distributive Property :

Ex 1 : If A = {1, 2, 3, 4, 5}, B = {1, 5, 6, 7, 8, 9} and C = {2, 3, 4, 5, 7}then find (i) A ∪ (B∩ C) and (ii) (A ∪ B) ∩ (A ∪ C). Show that both thesets are equal.

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Solution : (i) To find A ∪ (B∩ C)

(B∩ C) = {1, 5, 6, 7, 8, 9}∩ {2, 3, 4, 5, 7}

(B∩ C) = {5,7}

A ∪ (B ∩ C) = {1, 2, 3, 4, 5} ∪ {5, 7}

∴ A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 7} ............. (i)

(ii) To find (A ∪ B) ∩ (A ∪ C)

(A ∪ B) = {1, 2, 3, 4, 5} ∪ {1, 5, 6, 7, 8, 9}

(A ∪ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9}

(A ∪ C) = {1, 2, 3, 4, 5} ∪ {2, 3, 4, 5, 7}

(A ∪ C) = {1, 2, 3, 4, 5, 7}

(A ∪ B)∩ (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} ∩ {1, 2, 3, 4, 5, 7}

(A ∪ B)∩ (A ∪ C) = {1, 2, 3, 4, 5, 7} .............. (ii)

From (i) and (ii) A ∪ (B∩ C) = (A ∪ B)∩ (A ∪ C)

∴ Union of sets is distributive over Intersection of sets

Ex 2 : If A = {5, 10, 15, 20}, B = {5, 15, 20, 30, 35} and C = {5, 10, 25}then find (i) A ∩ (B ∪ C) and (ii) (A∩ B) ∪ (A ∩ C). Show that A∩ (B ∪ C)= (A∩ B) ∪ (A ∩ C).

Solution : (i) To find A ∩ (B ∪ C)

B ∪ C = {5, 15, 20, 30, 35} ∪ {5, 10, 45}

B ∪ C = {5, 10, 15, 20, 30, 35, 45}

∴ A∩ (B ∪ C) = {5, 10, 15, 20} ∩ {5, 10, 15, 20, 30, 35, 45}

∴ A∩ (B ∪ C) = {5, 10, 15, 20}................. (i)

(ii) To find (A∩ B) ∪ (A ∩ C)

A∩ B = {5, 10, 15, 20} ∩ {5, 15, 20, 30, 35}

A∩ B = {5, 15, 20}

A∩ C = {5, 10, 15, 20}∩ {5, 10, 25}

A∩ C = {5, 10}

∴ (A ∩ B) ∪ (A ∩ C) = {5, 10, 15, 20}.................. (ii)

From (i) and (ii) A∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

∴ Intersection of sets is distributive over Union of sets

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2. De Morgan’s LawsIf U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {0, 2, 4, 6, 8} and B = {0, 3, 6, 9}

(I) Find a) A| b) B| c) (A ∪ B)| d) (A ∩ B)|

(II) Verify a) (A ∪ B)| = A| ∩ B| b) (A ∩ B)| = A| ∪ B|

(I) a)To find A|

A| = U – A [The Set contains elements of U which arenot the elements of A ]

A| = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {0, 2, 4, 6, 8}

∴ A| = {1, 3, 5, 7, 9}.............. (i)

b) To find B|

B| = U – B [The Set contains elements of U which arenot the elements of B]

B| = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {0, 3, 6, 9}

B| = {1, 2, 4, 5, 7, 8}............... (ii)

c) To find (A ∪ B)|

A ∪ B = {0, 2, 4, 6, 8} ∪ {0, 3, 6, 9}

= {0, 2, 3, 4, 6, 8, 9}

(A ∪ B)| = U – (A ∪ B)

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {0, 2, 3, 4, 6, 8, 9}

∴ (A ∪ B)| = {1, 5, 7} .............. (iii)

d) To find (A∩ B)|

A∩ B = {0, 2, 4, 6, 8} ∩ {0, 3, 6, 9}

A∩ B = {0, 6}

(A∩ B)| = U – (A∩ B)

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {0, 6}

(A∩ B)| = {1, 2, 3, 4, 5, 7, 8, 9} .............. (iv)

(II) a) Verification of (A ∪ B)| = A| ∩ B|

From (i) and (ii), we getA| ∩ B| = {1, 3, 5, 7, 9}∩ {1, 2, 4, 5, 7, 8}

∴ A| ∩ B| = {1, 5, 7} .................. (v)From (iii) and (v) (A ∪ B)| = A| ∩ B|

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b) Verification of (A ∩ B)| = A| ∪ B|

A| ∪ B| = {1, 3, 5, 7, 9} ∪ {1, 2, 4, 5, 7, 8}

∴ A| ∪ B| = {1, 2, 3, 4, 5, 7, 8, 9}............... (vi)

From (iv) and (vi) (A∩ B)| = A| ∪ B|

I (A ∪ B)| = A| ∩ B| The complement of Union of sets is theintersection of their complements

II (A∩ B)| = A| ∪ B| The complement of intersection of setsis the union of their complements

These two laws are named after De Morgan Augustus as De Morgan's Laws

3. Relation between the number of elements of two sets :Let A = {a, b, c, d, e} and B = {c, f, g, b, h}

Draw the Venn diagram for A ∪ B

(i) From the above diagram find how manyelements are in the set.

a) A b) B c) A ∪ B d) A∩ B

(ii) Check whether n(A) + n(B) = n(A ∪ B) + n(A ∩ B)

(i) There are 5 elements in set A and denoted by (a) n(A) = 5

Similarly, b) n(B) = 5

c) n(A ∪ B) = 8

d) n(A ∩ B) = 2

(ii) Verify that n(A) + n(B) = n(A ∪ B) + n(A ∩ B)

From (a) and (b),

n(A) + n(B) = 5 + 5 =10 ............... (i)

From (c) and (d),

n(A ∪ B) + n(A∩ B) = 8 + 2 = 10 ............... (ii)

From (i) and (ii),

n(A) + n(B) = n(A ∪ B) + n(A∩ B)

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Note : i) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

ii) n(A∩ B) = n(A) + n(B) – n(A ∪ B)

iii) If A and B are disjoint sets then A∩ B = �

∴ a) n(A ∩ B) = 0

b) n(A ∪ B) = n(A) + n(B)

Ex 1 : There are Seven passengers in a compartment of a train. Five can speak Hindi,Four can speak English and two can speak both the languages. Find how manypassengers can speak (i) only Hindi (ii) only Enlgish. Draw Venn diagram.

Solution : 1) Let A be the set of passengers who can speak Hindi∴ n(A) = 5

2) Let B be the set of passengers who can speak English∴ n(B) = 4

3) A∩ B is the set of all passengers who can speak both Englishand Hindi in the compartment∴ n(A∩ B) = 2

4) The number of passengers who can speak only Hindi is

= n(A) – n (A ∩ B)

= 5 – 2 = 3

5) The number of passengers who can speak only English is

= n(B) – n(A∩ B)

= 4 – 2 = 2

Ex 2 : There are 60 Students in a class, every student learns atleast one of the subjectsKannada or English. 45 students offer Kannada and 30 English. How manyStudents offer i) Both the Subjects ii) Only Kannada iii) Only English.Draw Venn diagram

Solution : 1) K is the set of students who offer Kannada ∴ n(K) = 45

2) E is the set of students who offer English. ∴ n(E) = 30

3) K ∪ E is the set of students in the class. ∴ n(K ∪ E) = 60

A B

223

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(i) The number of students who offer both the subjects = n(K∩ E)

We know that, n(K ∩ E) = n(K) + n(E) – n(K ∪ E)

= 45 + 30 – 60

n(K ∩ E) = 75 – 60 = 15

(ii) The number of Students who offer only Kannada = n(K) – n(K∩ E)

= 45 – 15

= 30

(iii) The number of students who offer only English = n(E) – n(K ∩ E)

= 30 – 15

= 15

Ex 3 : In a survey of 1000 persons in Bangalore, it was found that 800 read newspaper X, 300 read news paper Y, and 200 read both the news papers. Howmany persons do not read both ?

Given : i) U is the set of persons under survey ∴ n(U) = 1000

ii) A is the set of the persons who read newspaper X ∴ n(A) = 800

iii) B is the set of the persons who read newspaper Y ∴ n(B) = 300

iv) A∩ B is the set of persons who read both newspaper X and Y

∴ n (A ∩ B) = 200

To find : 1) A ∪ B is the set of persons who read either X or Y

2) (A ∪ B)| is the set of persons who read neither X nor Y

Solution: 1) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

= 800 + 300 – 200

= 900

∴ The number of persons who read one of the papers = 900

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2) n(A ∪ B)| = n[U– (A ∪ B)]

= n(U) – n(A ∪ B)

= 1000 – 900

= 100

∴ The number of persons who read neither X nor Y = 100

Exercise : 1.11) a) If A = {0, 2, 4, 6, 8,} and B = {x : x is an even digit less than 5}

Find i) A ∪ B ii) B ∪ A iii) A ∩ B iv) B∩ A

b) If A = {1, 2, 3, 4, 5} and B = {2, 4, 0, 8}Verify i) A ∪ B = B ∪ A ii) A ∩ B = B ∩ A

2) a) If A = {p, q, r, s}, B = {s, t, u} and C = {s, t, u, v, w}Find i) A ∪ (B ∪ C) ii) (A ∪ B) ∪ C iii) A ∩ (B ∩ C) iv) (A ∩ B) ∩ C

b) If A = {1, 4, 9, 16}, B = {3, 4, 5} and C = {3, 9, 12}Verify i) A ∪ (B ∪ C) = (A ∪ B) ∪ C ii) A ∩ (B ∩ C) = (A ∩ B) ∩ C

c) If A = {5, 7, 9}, B = {7, 9, 11} and C = {9, 11}Verify i) A ∪ (B ∪ C) = (A ∪ B) ∪ (A ∪ C)

ii) A ∩ (B ∩ C) = (A∩ B)∩ (A ∩ C)

3) a) If A = {1, 3, 4, 8, 9, 12}, B = {1, 4, 9} and C = {2, 4, 8, 10}

Find i) A ∪ (B ∩ C) ii) A∩ (B ∪ C) iii) (A ∪ B ) ∩ (A ∪ C)iv) (A∩ B ) ∪ (A∩ C)

b) If A = {2, 4, 6, 8, 10}, B = {1, 2, 3, 4, 5, 6} andC = {1, 3, 5, 7, 9, 11, 13}

Verify i) A ∪ (B∩ C) = (A ∪ B)∩ (A ∪ C)

ii) A ∩ (B ∪ C) = (A∩ B ) ∪ (A∩ C)

iii) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)

iv) (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

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c) If X = { x : x is a prime number less than 12}Y = { x : x is an even number less than 12}Z = { x : x is an odd number less than 12}

Show that i) Union of sets is distributive over Intersection of sets.ii) Intersection of sets is distributive over Union of sets.

4) If U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}A = { x : x is a perfect square less than 10}B = {x: x is a multiple of 3 less than 10}

Verify i) (A ∪ B)| = A| ∩ B|

ii) (A∩ B)| = A| ∪ B|

5) a) A Florist has certain number of garlands. 110 of them have champak, 50have jasmine flowers and 30 garlands have both the flowers. Find the totalnumber of garlands with him.

b) In a class of 60 students every one should select either Mathematics orScience or both. If 45 students select Mathematics, 30 students selectScience, how many students did select both the subjects ?

c) In a survey of 1000 families of T.V. viewers in a city, it was found that750 families view Kannada programmes, 400 families view Hindi programmesand 300 families view both the programmes. Find how many families doview i) Kannada programmes only ii) Hindi programmes only iii) Neitherof the programmes.

SEQUENCE

1. Definition of a sequence :In our daily life we come across the arrangement of objects or numbers in an

order such as arrangement of students in a row as per their roll numbers, arrangementof books in the library, etc.

Arrange the books according to assigned number .

Number 1 2 3 4 5

Book D B C A E

Serial no. of book Book

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An Arrangement of numbers depends on the given rule.

Sl. No. Given Rule Arrangement of numbers

1 Write 4 and then add 5 successively 4, 9, 14, 19, 24 ................

2 Write 4 and then multiply by 5 4, 20, 100, ........................successively

3 Write 4 and then subtract .............................................3 successively

4 Write alternatively 1 and –1 ............................................

5 Write 2

1 and then add 1 to numerator

as well as denominator successively2

1,

3

2,

4

3, .............................

A Sequence is an ordered arrangement of numbers according to a given rule.

2. The terms of a sequence :

The individual numbers that form a sequence are the terms of a sequence.

Ex : The numbers 1, 3, 5, 7, ……… form a sequence

1, 3, 5, 7 ……… are called the first, second, third and fourth ………. termsof the sequence respectively. The terms of a sequence in successive order is denotedby T

n.

Order number of the terms 1st 2nd 3rd ……. nth …..…

Corresponding Symbols T1

T2

T3 …………..

Tn

……..

The nth term “Tn” is called the general term of the sequence.

3. Finite and Infinite sequence :1. A sequence having a finite number of terms is a finite sequence.

T1, T

2, T

3, T

4 ...........T

n is a finite sequence of n terms, here T

n is the last term.

Ex : i) 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 is the finite sequence of 10 terms.

ii) 5, 10, 20, 40 ………….. 1000 is also a finite sequence.

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2. A sequence having an infinite number of terms is an infinite sequence.

T1, T

2 , T

3, ………….. is an infinite sequence.

i) 3, 9, 27, 81 ……….. is an infinite sequence.

ii) 6, 4, 2, 0, -2 ………. is also an infinite sequence.

Worked Examples :

I. Write the nth term for the following sequences :

1) 5, 8, 11, 14 …………………..

Solution : We write first 5 and then add 3 successively.

T1

= 5,

T2

= 8 = 5 + 3

T3

= 11 = 5 + 6 = 5 + 2 x 3

T4

= 14 = 5 + 9 = 5 + 3 x 3∴ T

n= 5 + (n – 1) 3

2) 10, 100, 1000 …………………

Solution : We write natural powers of 10, successively.

T1

= 101

T2

= 100 = 102

T3

= 1000 = 103

∴ Tn

= 10n

3)10

1,

20

1,

30

1 ……………..

Solution : We write the reciprocals of multiples 10

T1

=10

1

T2

=20

1 =

2x10

1

T3

=30

1 =

3x10

1

Tn

=xn10

1 =

n10

1

13

4)2

1,

3

2,

4

3,

5

4 ………………..

Solution : We divide 1 by 1 + 1, 2 by 2 + 1, 3 by 3 + 1 and so on

T1

=11

1

+ = 2

1

T2

=12

2

+ = 3

2

T3

=13

3

+ = 4

3

∴ Tn

=1n

n

+

We describe the sequence by listing the first few terms (at least three terms) ofthe sequence or defining an algebraic expression for the general term T

n, then to find

the succeeding terms assign the values to n, as 1, 2, 3, 4 ……………… We denotesequence as {T

n}.

II. Worked Examples :1) Find the first four terms of the sequence whose nth term is 5n + 3.

Solution : Here Tn

= 5n + 3

To find T1, substitute n = 1 in T

n

T1

= 5 x 1 + 3

= 5 + 3 = 8

To find T2, substitute n = 2 in T

n

T2

= 5 x 2 + 3

= 10 + 3 = 13

To find T3, substitute n = 3 in T

n

T3

= 5 x 3 + 3

= 15 + 3 = 18

Similarly, T4

= 5 x 4 +3

= 20 + 3 = 23

The first four terms of the sequence are 8, 13, 18, 23.

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2) If Tn

= n2 +1, find the first four terms of the sequence.

Solution : Tn

= n2 +1

∴ T1

= 12 + 1 = 1 + 1 = 2

T2

= 22 + 1 = 4 + 1 = 5

T3

= 32 + 1 = 9 + 1 = 10

T4

= 42 + 1 = 16 + 1 = 17

The first four terms of the sequence are 2, 5, 10 and 17.

3) If Tn

= 4n + 1 find a) Tn-1

b) Tn+1

Solution : Here Tn

= 4n + 1

a) To find Tn-1

, Substitute n – 1 in the place of n

∴ Tn-1

= 4(n – 1) + 1

= 4n – 4 + 1

∴ Tn-1

= 4n – 3

b) To find Tn+1

, Substitute n + 1 in the place of n

Tn+1

= 4(n + 1) + 1

∴ Tn+1

= 4n + 4 + 1

∴ Tn+1

= 4n + 5

4) In a sequence, Tn

= 4n2 – 1. Find the value of n so that Tn

= 35

Solution : Tn

= 4n2 – 1

Tn

= 35 is given

∴ 4n2 – 1 = 35

4n2 = 35 + 1 = 36

∴ n2 =4

36 = 9

∴ n = 9 = ± 3

∴ ‘n’ is always a natural number. Therefore n = 3

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4. SeriesThe sum of terms of a sequence is called the series of the corresponding sequence.

T1

+ T2

+ T3

+ ……….. is an infinite series, whereas

T1

+ T2

+ T3

+ ………… + Tn is a finite series of n terms.

Usually the series of finite number of n terms is denoted by Sn or simply S.

(i.e) Sn

= T1

+ T2

+ T3

………… + Tn

Worked Examples :1) In the series, T

n= 3n + 1. Find S

5

Solution : S5

= T1

+ T2

+ T3

+ T4

+ T5

Now Tn = 3n + 1

T1

= 3 (1) + 1 = 3 + 1 = 4

T2

= 3 (2) + 1 = 6 + 1 = 7

T3

= 3 (3) + 1 = 9 + 1 = 10

T4

= 3 (4) + 1 = 12 + 1 = 13

T5

= 3 (5) + 1 = 15 + 1 = 16

∴ S5 = T

1 + T

2 + T

3 + T

4 + T

5

= 4 + 7 + 10 + 13 + 16

∴ S5 = 50

2) In a series Tn

= xn-1 (x ≠ 0) Write the infinte series.

Solution : Here Tn

= xn-1

Find the first four terms.

∴ T1 = x1-1

= x0 = 1

T2 = x2-1 = x1

T3 = x3-1 = x2

T4

= x4-1 = x3

The infinite series is 1 + x1 + x2 + x3 + ....................

Think! S

n – S

n-1 = T

n

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Exercise : 1.2

1) Write the next four terms of the following sequences given.

a) 5, 7, 9, 11 ................... b) 1, 2

1,

4

1,

8

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

c) 1

2,

2

3,

3

4,

4

5 ................... d) 4, -4, 4, -4 ..................

2) Write the first four terms of the sequence whose nth term isa) T

n = 2n-1 b) T

n = 5 - 2n c) T

n = 3n2 + 1 d) T

n = 3n-1

3) (a) If Tn = n find i) T

n+1 ii) T

n-1

(b) If Tn = n2 –1 find i) T

n–2 ii) T

n+1

(c) If Tn = 2n2 + 1 find the value of n if T

n = 73

(d) In a sequence Tn

= 5 – 3n find i) Tn+1

ii) Tn+2

4) In a series (i) Tn = 2n – 1 find S

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If (ii) Tn = (–1)n then find a) S

1b) S

2 c) S

3d) S

4

Hence Prove that S1

= S3

and S2

= S4

ARITHMETIC PROGRESSION

1. The definition of an arithmetic progression :We have already studied about the sequence. Now let us discuss the particular

kind of sequence which has a constant relationship between a pair of successive terms.

Consider the sequence1) 10, 12, 14, 16 ………….

2) 8, 5, 2, –1, –4 ……………

In Ex : 1, Let us calculate T2

– T1

, T3

– T2

, T4

– T3

Similarly we have T2

– T1

= T

3 – T

2 = T

4 – T

3 =

………….. = 2

In Ex : 2, T2

– T1

= T

3 – T

2 = T

4 – T

3 =

.............. = – 3

Such kind of sequences are called arithmetic progressions.

17

Definition :- An Arithmetic Progression is a sequence in which the difference between

a term and it’s preceding term is a constant. The constant is the common difference

(c.d) and is denoted by ‘d’.

(The Arithmetic Progression is abbreviated as A.P).

2. The general form and term of an A.P :

T1, T

2,

T3

………….. Tn .......... be an A.P

T2

–T1

= T3

–T2

= T

4 –T

3 =

………….. = T

n – T

n-1 = d

Let T1

= a = a + (1 – 1) d

∴ T2

= T1

+ d = a + d = a + (2 – 1)d

T3

= T2

+ d = a + 2d = a + (3 – 1)d

T

4 = T

3 + d = a + 3d = a + (4 – 1)d

………………………………….

Tn

= Tn-1

+ d = a + (n – 1)d

∴ a, a + d, a + 2d, a + 3d, ...................., a + (n – 1)d is the general form of an

arithmetic progression with ‘a’ as the first term and having common difference ‘d’. This

is the Standard form of an A.P. The last term a + (n – 1) d is denoted by Tn or l

Note : Tn

= a + (n – 1)d is the general term of an A.P

Note : 1. In an A.P succeeding term of a given term is obtained by adding ‘d’

to it. [Tn+1

= Tn+ d].

2. In an A.P preceding term of a given term is obtained by subtracting

‘d’ from it [Tn-1

= Tn

– d].

Worked Examples :1) The first term of an A.P. is 3 and common difference is 5. Find the A.P.

Solution : a = 3, d = 5, Write the A.P.

T1 = a = 3

T2 = T

1+ d = 3 + 5 = 8

T3

= T2

+ 5 = 8 + 5 = 13

T4

= T3

+ 5 = 13 + 5 = 18

A.P. is 3, 8, 13, 18, ......................

18

2) Find the nth and the 20th terms of the A.P. 3, 7, 11, 15, ...............

Solution : a = 3, d = T2

– T1

= 7 – 3 = 4, Tn

= ?, T20

= ?

Tn

= a + (n – 1)d.

Tn

= 3 + (n – 1)4

= 3 + 4n – 4

Tn

= 4n – 1

∴ T20

= 4(20) – 1 (OR) T20

= 30 + (20 – 1)4

= 80 – 1 = 3 + 19 x 4

T20

= 79 T20

= 3 + 76 = 79

3) In an A.P prove that T10

= T6

+ 4d.

Solution : Tn = a + (n – 1)d.

L.H.S = T10

= a + (10 – 1)d

L.H.S = a + 9d .................... (i)

R.H.S = T6

+ 4d

= a + (6 – 1)d + 4d

= a + 5d + 4d

R.H.S = a + 9d ................. (ii)

From (i) and (ii)

T10

= T6

+ 4d

In general we can prove that Tp

= Tq

+ (p–q)d

Note : i) d = )qp(

)TT( qp

−−

ii) In particular, d = )1n(

)aT( n

−−

4) The 20th and 30th terms of an A.P. are 201 and 301 respectively. Find c.d.

Solution : Tp

= T30

= 301, Tq

= T20

= 201

∴ d =)qp(

)TT( qp

−−

∴ d = )2030(

)201301(

−−

= 10

100 = 10 ∴ d = 10

19

5) How many terms are there in the A.P. 4, –1, –6, ............., (–106)

Solution : Given : a = 4, d = –5, Tn = –106, n = ?

Tn = a + (n – 1)d.

∴ a + (n – 1)d = Tn

4 + (n – 1) (–5) = –106

4 – 5n + 5 = –106

–5n + 9 = –106

∴ –5n = –106 – 9

∴ –5n = –115

n = 5

115

−−

= 23 ∴ n = 23

There are 23 terms in the given A.P.

6) The fourth and eighth terms of an A.P. are in the ratio 1 : 2 and tenth term is 30,Find the Common difference and the first term. Write the A.P.Solution : It is given that T

4 : T

8 = 1 : 2 and T

10 = 30

8

4

T

T =

2

1

∴ 2T4

= T8

2(T10

– 6d) = T10

– 2d

∴

T10

= T4+ 6d

2(30 – 6d) = 30 – 2d and T10

= T8

+ 2d

60 – 12d = 30 – 2d

–12d + 2d = 30 – 60

–10d = –30

∴ d = 10

30

−−

= 3 ∴ d = 3

To find ‘a’

a + (n –1)d = Tn

a + (10 –1)3 = 30

a + 9 x 3 = 30

a + 27 = 30

∴ a = 30 – 27 = 3 ∴ a = 3

The A.P. is 3, 3 + 3, 3 + (2 x 3), 3 + (3 x 3), ....................

i.e 3, 6, 9, 12, .....................

20

7) The angles of a triangle are in A.P. The smallest angle is 300. Show that the triangleis a right angled triangle.

Solution : In triangle ABC, let us say ∠ A = 300

Since ∠ A, ∠ B, and ∠ C are in A.P. ∠ B = 300+d and∠ C = 300 +2d

We know that ∠ A + ∠ B + ∠ C = 1800

∴ 300 + 300 + d + 300 + 2d = 1800

900 + 3d = 1800

∴ 3d = 1800 – 900

3d = 900

∴ d =3

900

= 300

∴ d = 300

The other two angles of the triangle are∠ B = 300 + d and ∠ C = 300 + 2d

= 300 + 300 = 600 = 300 + 2(300) = 900

∴ The triangle ABC is a right angled triangle.

8) Find the three numbers of an A.P. whose sum is 12 and their product is 48.

Solution : Let ‘a’ be the middle number.

Since three numbers are in A.P. the other two numbers are a – d and a + d.

∴ (a – d) + a + (a + d) = 12 ∴ 3a = 12

a = 3

12 = 4

∴ a = 4(a – d) a (a + d) = 48

∴ 4 (4 – d) (4 + d) = 48

(4 – d) (4 + d) =4

48 = 12

16 – d2 = 1216 – 12 = d2

d2 = 4

d = ± 4 = ± 2 ∴ d = ± 2

∴ The three numbers are 2, 4 and 6 (or) 6, 4 and 2

21

3. The sum of a finite arithmetic series :A series whose terms are in A.P is an arithmetic series

Ex : 1) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 is a finite arithmeticseries and its sum is 55.

2) 1 + 2 + 3 + ……….. + 1000 is an arithmetic series of 1000 termsand its sum is 5,00,500.

This can be calculated by actual adding or with the help of a calculator. But boththese methods are very time consuming. For this there are very simple and interestingformulae for the sum of the first ‘n’ natural numbers and sum of the first ‘n’ termsof an A.P. (Arithmetic series)

I 1) Find the sum of the first ‘n’ natural numbers1, 2, 3, …………. (n–2), (n–1) and n are the first ‘n’ natural numbers

Solution : Sn = 1 + 2 + 3 + …………………+ (n – 2) + (n – 1) + n

+ Sn = n + (n – 1) + (n – 2) + ………………… + 3 + 2 + 1

2Sn = (n + 1) + (n + 1) + (n + 1) + ………. + (n + 1) + (n + 1)

∴ 2Sn = n(n + 1) [

∴

There are n terms in the R.H.S]

∴ Sn =

2)1n(n +

Usually the sum of the first ‘n’ natural numbers is denoted by ∑n or ∑n

1

n

1 + 2 + 3 + 4 + ………… + n = 2

)1n(nn

+=∑

1) 1 + 2 + 3 + …………. + 20 = ∑20

1

n

=2

)120(20 + = 10 x 21 = 210

2) 1 + 2 + 3 + .…....….+ 1000 = ∑1000

1

n

=2

)11000(1000 + = 500 x 1001 = 500500

Note : ∑ − )1n( =2

)11n)(1n( +−− =

2

n)1n( −

22

Try :If a is constant then

a + 2a + 3a + . . . . . + na = ?

Activity :

What is )1n(n −∑+∑ = ?

Count the number of rings and horizontal bars of thisrow. Count the vertical bars below or above this rowthen answer the following1. How many rings are there in the design ?2. How many horizontal bars are there in the design?3. How many vertical bars are there in the design ?4. If the radius of each ring 3.5 cm and length

of each bar is 9 cm. what is the length of thesteel bar used for the design ?

3) Find the sum of all positive multiples of 3 less than 50.

Solution : 3, 6, 9, ..………, 48 are all the positive multiples of 3 < 50.

∴ Sn

= 3 + 6 + 9 + . . . . . . . . . . . . . . . . + 48

= 3(1 + 2 + 3 + . . . . . . . . . . + 16)

= 3

∑

16

1

n

= 3 x 2

1716 ×

=

2

816 = 408

II Find the sum of first ‘n’ terms of an Arithmetic ProgressionSolution : Let a, a + d, a + 2d, .........., a +(n – 1)d are the n terms of an A.P.(i.e) S

n= a + (a + d) + (a + 2d) + .................. + [a + (n – 1)d]= (a + a + a + .........upto n terms) + [d + 2d + 3d + ..........+ (n – 1)d]= na + [1 + 2 + 3 + ..............+ (n – 1)]d

∴ Sn

= na +

−∑

−1n

1

)1n( d

= na +2

)11n)(1n( +−− x d

= 2

dxn)1n(na2 −+

23

Sn

=2

n [2a + (n – 1)d]

∴ Sn

=2

n [a + a + (n – 1)d]

∴ Sn

=2

n(a + l)

Worked Examples :1) In an A.P of 21 terms, the first and the last terms are 4 and 64 respectively. Find

the sum of the A.P.

Solution : n = 21, l = 64, a = 4

Sn

= 2

n (a + l)

S21

= 2

21(4 + 64)

S21

= 2

21 x 68 = 21 x 34 = 714

∴ S21

= 714

2) Find the sum of the first 25 terms of the series 3 + 7 + 11 + 15 + ............

Solution : Given Sn

= 2

n[2a + (n – 1)d]

S25

=2

25[2 x 3 + (25 – 1)4]

S25

=2

25(6 + 24 x 4)

S25

=2

25(6 + 96)

S25

=2

25x 102 = 25 x 51 = 1275 ∴ S

25 = 1275

Do You Know ?Sum of n terms of an A.P isn times the A.M of the firstand the last terms.

n = 25a = 3d = 4S

25 = ?

24

3) Find the sum of all natural numbers between 91 and 170 which are divisibleby 5.Solution : The numbers between 91 and 170 which are divisible by 5 are

95, 100, 105, .............. 165S

n= 95 + 100 + 105 + ............... + 165

Tn = 165 First find the number of terms in the A.P.

∴ a + (n – 1)d = 16595 + (n – 1)5 = 165

∴ 95 + 5n – 5 = 165

∴ 5n + 90 = 1655n = 165 – 905n = 75

∴ n =5

75 = 15

n = 15

∴ Sn

=2

n (a + l)

∴ S15

=2

15(95 + 165)

=2

15x 260

= 15 x 130

∴ S15

= 1950

4) The sum of n terms of an arithmetic series is Sn = 2n2 + 6n. Find the first term

and the common difference.Solution : Given : S

n= 2n2 + 6n and S

1 = T

1

∴ S1

= 2(1)2 + 6(1)

= 2 + 6 = 8

T1

= 8 ∴ a = 8

To find 'd' S2

= 2(2)2 + 6(2)

= 2(4) + 12

= 8 + 12 ∴ S2

= 20

S2

= T1

+ T2

∴ T

1+ T

2= 20

a = 95d = 5l = T

n = 165

25

8 + T2

= 20

T2

= 20 – 8 = 12

T2

= 12 ∴ d = T

2– T

1

= 12 – 8 ∴ d = 4

5) How many terms of the series 10 + 8 + 6 + ............ should be added to get thesum –126.Solution : S

n = –126

a = 10, d = –2, Sn = –126

∴ 2

n[2a + (n – 1)d] = –126

∴ 2

n[2 x 10 + (n – 1)x –2] = –126

2

n[20 – 2n + 2] = –126

2

n[22 – 2n] = –126

∴ n(11 – n) = –126 ∴ 11n – n2 = –126

(i.e) – n2 + 11n + 126 = 0

(i.e) ∴ n2 – 11n – 126 = 0

n2 – 18n + 7n – 126 = 0

n(n – 18) + 7 (n – 18) = 0

(n – 18) (n + 7) = 0

n – 18 = 0 or n + 7 = 0

n = 18 or n = –7

As n is a positive integer n = 18

6) Sanganbasava rides a bicycle from his home to the ashram. He covers 125 metersat the end of first minute, 135 meters at the end of second minute and so on.If he reaches the Ashram at the end of 10 minutes. Find the distance between hishome and ashram.

Solution : We can observe that the distances covered by Sanganbasava insuccessive minutes are in A.P.

The required distance = S10

= 125 + 135 + 145 + ............... upto 10 terms.

a = 10

d = –2

Sn = –126

26

a = 125, d = 10, n = 10

Sn

=2

n[2a + (n – 1)d]

∴ S10

=2

10[2 x 125 + (10 – 1)10]

= 5[250 + 9 x 10]

= 5(250 + 90)

= 5 x 340

= 1700 meters

S10

= 1.7 Kilometers

∴ The distance of the Ashram from his house is 1.7 Kms.

Exercise : 1.31) Which of the following sequences are in A.P.

a) 3, 7, 11, 15 ............... b) 8, 4, 2, 1, ................... c) 1, –1, –3, –5, .........

2) Find :a) T

n and T

15 for the A.P 1, 4, 7, 10, ..............

b) T20

for the A.P 8, 2, – 4, –10, ...................

3) A. Find a) ∑25

1

n b) ∑∑ +25

1

30

1

nn c) ∑∑ −30

1

50

1

nn

B. a) If ∑n = 55 find n b) If ∑n = 210 find n.

C. Find the sum of all natural numbers below 50 which are divisible by 7.

4) In an A.PA. a) If a = 11, d = 2, n = 10 find T

10and S

10

b) If a = 12, d = 3, Tn

= 84 find n

c) If d = 4 and T35

= 123 find a

d) If a = 3 and T20

= 60 find d

B. i) How many terms are there in A.P 4, –1, –6, ..........., (–106)

ii) Test whether 25 is a term of the A.P 3, 9, 12, 15, ...........

iii) Test whether 36 is a term of the A.P 4, 8, 12, 16, ..............

27

5) A. Find the A.P in which a)Tn = 2n + 1 b) T

n= 4 – 5n c) S

n= 5n2 + 3n

B. a) In an A.P. the fourth and seventh terms are 17 and 23 respectively, find‘d’ and ‘a’

b) In an A.P. T10

= 20 and T20

= 10 find T30

c) The fifth and tenth terms are in the ratio 1 : 2 and T12

= 36, find A.P.

6) A. Find the sum of the following Arithmetic series.

1) 2 + 5 + 8 + .......... upto 12 terms

2) 1 + 4 + 7 + ........... upto 22 terms

3) 3 + 7 + 11 + ........... upto 35 terms

4) Find the sum of the Arithmetic series which contains 25 terms and whosemiddle term is 20.

B. In an A.P. If 1) Tn

= 4n + 3 find S15

2) Tn = 5 – 2n find S

20

C. 1) Find the sum of all the first ‘n’ odd natural numbers.

2) Find the sum of all natural numbers between 101 and 201 which are divisibleby 4.

7) A) Find the sum of the finite series

a) 1 + 6 + 11 + 16 + .......... + 96 b) 1 + 4 + 7 + ........... +73.

B) Find the number of terms in the following series if

a) 3 + 5 + 7 +..........= 624 b) 15 + 12 + 9 + 6 +...........= –90

8) a) Anup, Azhar and Anthony have 105 marbles altogether. Anup has 40 marblesIf their shares are in A.P., find the number of marbles with the other two boys.

b) The angles of a quadrilateral are in A.P. If the smallest angle is 150, find theangles of the quadrilateral.

c) Veershree climbed 23 steps of Golgumbaz in the first minute. After that sheclimbed two steps less than what she had climbed in the previous minute. Ifshe reached the whispering gallery of Golgumbaz after 7 minutes how manysteps did she climb to reach the whispering gallery?

d) Find the three numbers in A.P. whose sum and products are

i) 6 and 6 ii) 15 and 18 iii) 15 and 1050

28

GEOMETRIC PROGRESSION

1. The definition of a Geometric Progression :In the previous unit we have studied about the arithmetic progression in which

the difference between a term and its preceding term was constant.

Consider the following sequences :

1) 2, 4, 8, 16, ................ 2) 12, 6, 3, 2

3, ................

Calculate 1

2

T

T,

2

3

T

T,

3

4

T

T, ...................

Then we have 1

2

T

T =

2

3

T

T =

3

4

T

T = ....................= the ratio is constant.

Such sequences are Geometric Progressions.

Definition : A Geometric progression is a sequence in which the ratio of a termand it’s preceding term is a constant. This constant is called the common ratio and isdenoted by “r” (Geometric progression is abbreviated as G.P).

2. The General form and the general term of a Geometric progression :Let T

1, T

2, T

3, ............. T

n be a G.P in which T

1 = a and common ratio = r

Then 1

2

T

T =

2

3

T

T =

3

4

T

T = ................. = r

T1 = a = ar1-1

T2 = T

1 x r = ar = ar2-1

T3 = T

2 x r = (ar)(r) = ar2 = ar3-1

T4 = T

3 x r = ar2 x r = ar3 = ar4-1

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

Tn = T

n-1 x r = arn-1 ∴ T

n = arn-1

a, ar, ar2, ......................ar n-1 is a Geometric Progression with the first term ‘a’and common ratio r. This is the standard form of a G.P.

Activity : Take a rope of any length and divide it into 81 equal parts by foldingit in different stages. Observe

1. The number of equal folds at each stage.2. Find the length of each part at each stage.

Tn = arn-1 is the general term of the G.P.

29

Note : 1. To Obtain the Succeeding term of a given term Tn multiply it

by r.

Tn+1

= Tn x r

2. To Obtain the Preceding term of a given term Tn divide it by r.

Tn-1

= r

T n

Worked Examples :1) The first term and common ratio of a G.P are 3 and 4 respectively, find the G.P.

Solution : a = 3, r = 4∴ T

1= a = 3

T2

= ar = 3 x 4 = 12T

3 = ar2 = 3 x 42 = 3 x 16 = 48

T4 = ar3 = 3 x 43 = 3 x 64 = 192

∴ The G.P is 3, 12, 48, 192, ..............

2) In a sequence Tn = 2(1-n), Prove that the sequence is a G.P.

Solution : Tn

= 21-n

∴ T1

= 21-1 = 20 = 1

T2

= 21-2 = 2-1 =2

1

T3

= 21-3 = 2-2 =22

1 =

4

1

T4

= 21-4 = 2-3 =32

1 =

8

1 and so on …

1

2

T

T=

121

=2

1=

2

3

T

T=

2141

= 4

1 x

1

2 =

2

1

∴1

2

T

T=

2

3

T

T=

2

1 = r ∴ The given sequence is a G.P.

3) Find the nth and 7th term of the G.P 1, 3, 9, 27, ............Solution : a = 1, r = 3

Tn = arn-1

Tn

= 1.(3)n-1

Tn

= 3(n-1)

T7

= 37-1 = 36 ∴ T7 = 729

30

4) Which term of the G.P 3, 6, 12, 24, ............ is 96?

Solution : a = 3, r = 2, Tn = 96, n = ?

Tn = arn-1

i.e., arn-1 = Tn

3.2 n-1 = 96

2n-1 =3

96 = 32

∴ 2n-1 = 25

∴ n – 1 = 5

n = 5 + 1 = 6

Hence the sixth term is 96.

5) In a G.P seventh term is eight times the fourth term and the fifth term is 12, findthe G.P.

Solution : T7

= 8 x T4

∴4

7

T

T=

1

8 ∴

q

p

T

T = rp-q

3ar

ar6

=1

8

r3 = 8

r3 = 8 ∴ r = 2

To find ‘a’ : T5

= 12

∴ ar4 = 12

a x 24 = 12

a x16 = 12

a =16

12 =

4

3 ∴ a =

4

3

The required G.P is 4

3,

2x4

3,

22x

4

3,

32x

4

3 ..............

i.e. 4

3,

2

3, 3, 6 .............

( )

31

Note :

1) 1 = 1 = 1 2) 010

1= 1= 1

1.0

1=

1011

= 10 110

1=

10

1= 0.1

01.0

1=

10011

=100 210

1=

100

1= 0.01

001.0

1=

100011

=1000 310

1=

1000

1= 0.001

In Note 1) Here we observe that as the denominator decreases the number increases.We conclude that when the denominator becomes nearer and nearer tozero, the number will become very large which is beyond our imagination

and this is infinity and is denoted by ‘ ∞ ’ (Infinity). i.e 0

1 = ∞ .

In Note 2) Here we observe that as the power of 10

1 increases the number

decreases. We conclude that when the power n of 10

1 is very large

quantity ( ∞ ) then n10

1 will be nearer and nearer to zero. In general

if r < 1 then rn = 0 when n is nearer to ∞ .

3. To find the sum of n terms of a Geometric Progression with the first terma and the common ratio r.Let a, ar, ar2, ...............ar n-2, ar n-1 be the n terms of a G.P.

∴ Sn

= a + ar + ar2 + ar3 + .................+ arn-2 + arn–1

r. Sn

= ar + ar2 + ar3 + ar4 + ............... arn-1 + arn

∴ Sn

– rSn

= a – arn

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32

∴ (1 – r) Sn

= a (1 – rn)

∴ Sn

=)r1(

)r1(a n

−−

when r < 1

This formula can be written as Sn

=)1r(

)1r(a n

−−−−

∴ Sn =

)1r(

)1r(a n

−−

when r > 1

4. Sum of infinite terms of a G.P.We know that

Sn =

)r1(

)r1(a n

−−

when r <1 and if r < 1, rn approaches zero as n approaches ∞

∴ ∞S = )r1(

)01(a

−−

When ‘n’ approaches ‘ ∞ ’ here we write ∞S as

∴ ∞S = )r1(

a

−

Worked Examples :

1) Find the sum of 2 + 4 + 8 + ............. to 10 terms.

Solution : a = 2, r = 2

4 =

4

8 = 2 > 1, n = 10

∴ Sn =

)1r(

)1r(a n

−−

∴ S10

= )1r(

)1r(a n

−−

= )12(

)12(2 10

−−

S10

= 2(1024 – 1)

S10

= 2(1023)

∴ S10

= 2046

33

2) Find the sum of first 6 terms and ∞S terms of the G.P 1, 3

1,

9

1,

27

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

Solution : a = 1, r = 3

1, n = 6

(i) Sn

=)r1(

)r1(a n

−−

r < 1

S6

=)r1(

)r1(a 6

−−

= 1

−

−

31

1

31

16

=

3231

1 6−

=

−

729

11

2

3

=

−

729

1729

2

3=

729

728x

2

3

∴ S6

=243

364

(ii) 2

3

321

31

1

1

)r1(

aS =

=−

=−

=∞

2

3S =∞

3) Find the number of terms in the series 1 + 4 + 16 + .................. if the sum of theseries is 341.

Solution : a =1, r = 4, Sn

= 341

∴

r > 1, )1r(

)1r(aS

n

n −−=

∴ n

n

S)1r(

)1r(a =−−

34

341)14(

)14(1 n

=−−

3413

14n

=−

4n –1 = 341 x 3

4n = 1023 + 1

4n = 1024

4n = 45

∴ n = 5 The Series has 5 terms

4) Find the ratio of sum of the first four terms to the first eight terms of the G.P.

Solution : Sn

=)1r(

)1r(a n

−−

∴ S4

= )1r(

)1r(a 4

−−

......... (1)

S8

=)1r(

)1r(a 8

−−

........... (2)

Divide (1) by (2)

8

4

S

S=

)1r()1r(a

)1r()1r(a

8

4

−−

−−

= 2248

4

)1()r(

)1r(

)1r(

)1r( 4

−−=

−−

= )1r)(1r(

)1r(44

4

+−−

∴ S4

: S8 = 1 : (r4 + 1)

Note : i) S8

: S4 = r4 + 1 ii) S

2n ÷Sn = rn + 1

35

5) In a G.P. a = 3, r = 2 and Tn = 96 Show that S

n = 189.

Solution : a = 3, r = 2, Tn

= 96, Sn

= ?

Sn

=)1r(

)1r(a n

−−

=)1r(

aarn

−−

=1r

aT 1n

−−+

=12

32x96

−−

[

∴

Tn+1

= Tn x r]

=1

3192 −

∴ Sn

= 189

OR

arn-1 = Tn

3(2)n-1 = 96

2n-1 =3

96 = 32

2n-1 = 25

n – 1 = 5 ∴ n = 5 + 1 = 6 ∴ n = 6

To find S6 :

Sn

=)1r(

)1r(a 6

−−

S6

=)12(

)12(3 6

−−

=1

)164(3 −

= 3 x 63

∴ S6 = 189

36

Exercise : 1.4

1) Which of the following sequences are in G.P?

a) 5, 10, 20, ......................... b) 1, –1, 1, –1, ...........................

c) 2, 4, 6, 8, ........................ d) 1, 4, 9, 16, .............................

2) Find a) Tn and T

8 for 1, 2, 4, 8, .............

b) T6 for 1,

3

1− , 9

1,

27

1− ........ c) T9 for 2, 0.2, 0.02, 0.002, .......

3) Find the G.P if

1) a) Tn

= 2 x 5(n-1) b) Tn

= 2 x 3(n-1) c) T7

= 243 and r = 3

d) ∞S =3

2 and a = 1 e) ∞S = 3 and r =

3

1

2) In a G.P.

a) The first term is 3

1 and sixth term is

729

1, find r.

b) If T8

= 32 and r = 2, find a.

c) Which term of the G.P 1, 2, 4 ........................ will be 256.

4) I. Find the sum of the following Geometric series

a) 1 + 2 + 4 + ...................... upto 9 terms

b) 6 + 3 +2

3+ .......................upto 8 terms

c) 3 + 0.3 + 0.03 + 0.003 + .......................... upto 6 terms

d) 1 +3

1+

9

1+ ................................ upto ∞

II. a) Find the first term of a G.P in which S6

= 1365 and r = 4

b) Find the common ratio of a G.P in which S2

= 6 and a = 2

5) Find the sum of the following finite series

a) 1 + 2 + 4 + .............. + 256. b) 3

1+

9

1+ ..................... 103

1

c) 1 + 0.1 + 0.01 + ................... + (0.1)9

37

6) How many terms of the series

a) 2 + 4 + 8 + ................... make the sum 1022

b) 3 + 9 + 27 + ................. make the sum 1092

c) 2 + 6 + 18 + ................. make the sum 728

7) I. Find a) S3

: S6 for the series 4 + 12 + 36 + .................

b) S4

: S8 for the series 5 + 10 + 20 + ................

II. Find the G.P if

a) T3

: T6

= 1 : 8 and T5

= 64 b) T4

: T8

= 1 : 81 and T6

= 486

c) S10

: S5

= 33 : 1 and T6

= 32 d) S8

: S4

= 97 : 81 and T5

= 16

8) a) In a G.P of 6 terms the first and last terms are 5 and 160, find the remainingterms

b) Ganga marked a dot in the first square, 2 dots in the second, 4 in the thirdsquare and so on. How many dots should she mark in the 8th square?

c) Sahil purchased certain number of laddus. He gave half of them to the firstfriend, half of the remaining to the second and so on. He and His fifth friendate 33 laddus. How many laddus did Sahil purchase ?

HARMONIC PROGRESSION

1. The meaning of a Harmonic Progression :

Consider the sequence 5

1,

9

1,

13

1,

17

1, ................

Taking the reciprocal of each of the terms of the sequence.

We get 5, 9, 13, 17, ..................

Observe that this is again a sequence which is an A.P with a = 5 and d = 4

A Harmonic Progression (H.P) is a sequence in which the reciprocalof terms of it form an arithmetic progression.

38

2. The General term of a Harmonic Progression

If Tn is the nth term of a H.P then

nT

1 will be the nth term of the A.P

Since nth term of A.P = [a + (n – 1)d]

∴ nT

1 = a + (n – 1)d

∴ Tn

= d)1n(a

1

−+ is the general term of H.P

∴ a

1,

da

1

+ , d2a

1

+ , ....... d)1n(a

1

−+ is the standard form of a H.P.

Note :1) The reciprocal terms of a H.P form an A.P.

2) There is no formula to find the sum of n terms of a H.P as in the case ofA.P and G.P.

Worked Examples :

1) State which of the sequences are in H.P

a) 1, 2

1,

3

1,

4

1, ....... b) 1,

2

1,

4

1,

8

1, ....... c)

2

1,

4

1,

6

1, .........

Solution : Taking the reciprocal of every term of the given sequences.

a) 1, 2, 3, 4, ............ b) 1, 2, 4, 8, .......... c) 2, 4, 6, ...........

Now the reciprocal of a) and c) are in A.P

∴ They are Harmonic Progressions but in b) the reciprocals of the terms of thesequence are not in A.P ∴ It is not a H.P

2) Show that 6, 4, 3 ..................... are in H.P. Hence find the nth and 20th terms.

Solution : The Sequence of the reciprocals of given sequence is :

∴ 6

1,

4

1,

3

1, ................................

∴ T2

– T1 =

4

1 –

6

1T

3– T

2 =

3

1 –

4

1

= 12

23 − =

12

1 =

12

34 − =

12

1

39

∴ d = 12

1 and a =

6

1

∴ nth term of A.P = a + (n – 1)d

= 6

1+ (n – 1)

12

1

= 12

1n2 −+ ∴ T

n=

12

)1n( +

∴ nth term of H.P Tn

= d)1n(a

1

−+ =

+1n

12

∴ T20

of H.P =

+120

12 =

21

12

Example : 7th term of a H.P is 2

3 and its 10th term is

17

12 find T

8,

T7 and T

10 of A.P will be

3

2 and

12

17 respectively

d =)qp(

TT qp

−−

∴ d =

−

−

)107(1217

32

=

−

−

)3(12

178

=3

1x

12

9

−−

∴ d = 4

1

But T7 = a + 6d =

3

2

∴ a + 6 x 4

1 =

3

2

a + 2

3 =

3

2

40

a = 3

2 –

2

3 =

6

94 − =

6

5−

∴ a = 6

5−

T8 of A.P. =

−+−

4

1)18(

6

5 =

4

7

6

5 +−

= 12

2110 +−T

8 =

12

11

∴ T8 of H.P. =

11

12

Exercise : 1.5

1) Which of the following are Harmonic progressions.

a) 1, 3

1,

5

1,

7

1, ................. b) 1,

3

1,

6

1,

12

1, ...................

c) 1, 3

2,

2

1,

5

2, ..................... d)

2

1,

5

1,

10

1,

17

1, ................

2) Find

a) Tn in

2

1,

5

1,

8

1, .......

b) T12

in 8

1,

13

1,

18

1,

23

1, ...........

c) T10

in 5

1,

3

1, 1, -1, .................

3) a) In a H.P, T4

=12

1 and T

10=

42

1 find T

19

b) In a H.P, T5

=19

6 and T

1=

9

2 find T

7 and T

12

Think !

Can this problem besolved in an alternatemethod?

41

ARITHMETIC, GEOMETRIC AND HARMONIC MEANS BETWEENTWO NUMBERS

I. Arithmetic Mean(A.M) :If a, A, b are in A.P, then A is called the Arithmetic Mean between a and b.Since a, A and b are in A.P ∴ A – a = b – A

∴ A + A = a + b 2A = a + b

A =2

ba + is the formula to find the A.M between two numbers a and b.

Worked Examples :1) Find the A.M between 5 and 17.

Solution : A = 2

ba +

A = 112

22

2

175 ==+∴ A = 11

2) If 2x + 3, x – 1 and 2 – x are in A.P. find x.

Solution : a = 2x + 3, A = x – 1, b = 2 – x

∴ A =2

ba +

x – 1 =2

232 xx −++

1

1−x=

2

5+x

∴ 2x – 2 = x + 5 2x – x = 5 + 2 ∴ x = 7

II. Geometric Mean between two numbers (G.M) :If a, G, b are in G.P , then G is called the Geometric Mean between a and b.Since a, G and b are in G.P.

a

G =

G

b

G2 = ab

G = ab is the formula to find the G.M between two numbers a and b.

42

Worked Examples :1) Find the G.M between 5 and 16.

Solution : G = ab

= )16x5(

G = 54

2) If 4, x + 2 and 9 are in G.P find x.

Solution : a = 4, G = x + 2, b = 9

Since a, G, b are in G.P

∴ G = ab

G = 9x4

x + 2 = 36±

∴ x + 2 = 6±x + 2 = + 6 or x + 2 = – 6

x = 6 – 2 or x = –6 – 2

∴ x = 4 or x = –8

III. Harmonic Mean between two numbers (H.M) :If a, H, b are in H.P, then H is called the Harmonic Mean between a and b.

Since a, H and b are in H.P

a

1,

H

1,

b

1 are in A.P

∴ H

1–

a

1=

b

1–

H

1

∴ H

1+

H

1=

a

1+

b

1

∴ H

2=

ab

ba +∴ H(a+b) = 2ab

H = ba

ab2

+ This is the formula to find the H.M between two numbers a and b.

43

Worked Examples :

1) Find the H.M between 6 and 3.

Solution : H =ba

ab2

+

∴ H = 36

3x6x2

+

H =9

36∴ H = 4

IV. Relationship among A.M, G.M and H.M

Theorem : If A,G,H are the A.M, G.M and H.M of two positive numbers

a and b, then show that A,G and H are in G.P(G = AH )

Now A is A.M between a and b A =2

ba + ...............(1)

G is G.M between a and b G = ab .............. (2)

H is H.M between a and b H =ba

ab2

+ ............... (3)

G = ab

∴ G2 = ( ) 2

ab = ab

A.H =

+

+

ba

ab2

2

ba= ab

G2 = AH or G = AH

Thus G is the Geometric mean between A and H.

∴ A, G, H are in G.P

Note : A.M, G.M, and H.M between 4 and 16 are 10, 8 and 6.4respectively.

Now 10 > 8 > 6.4

In general A ≥ G ≥ H

a = 6b = 3

44

Exercise : 1.6

a) Find x if the following numbers are in A.P

1) 4, x, 12 2) 2, (x – 1), 4 3) 6 – a, x, 6 + a

b) Find x if the following numbers are in G.P

1) 16, x, 25 2) 5, x + 1, 20

c) If A, G, H are A.M, G.M, and H.M of the following pair of numbers

a) 1 and 4 b) 2 and 8 c) 9 and 16

Verify that i) A, G, H are in G.P ii) A > G > H.

MATRICES

Recall : A matrix is a rectangular arrangement of numbers in terms of rows and columnsenclosed within brackets.

Example : Matrix Order of Matrix

(i) A =

321

413(2x3) matrix

(ii) B =

61

34

25

(3x2) matrix

(iii) C =

987

654

321

(3x3) matrix

The Horizontal arrangements are called 'Rows' and the vertical arrangements arecalled 'Columns'. In example (i) matrix A has 2 rows and 3 columns. Hence its orderis 2x3.

1. Multiplication of a matrix by a scalar

Consider a (2x3) matrix A =

654

321 Multiply both sides by 2, we get

2A = 2

654

321

45

=

6x25x24x2

3x22x21x2

∴ 2A =

12108

642

Now Consider a (2x2) matrix

A =

34

25

Multiply both sides by K

∴ KA = K

34

25

KA =

K3K4

K2K5

2. Transpose of a matrixConsider a matrix “A” of order (2x3)

A =

231

654

Obtain a matrix by changing the rows into corresponding columns of A.

Then the transposed matrix is

26

35

14

This matrix is denoted by A|.

A| is called the transpose of the matrix A.

Example : A =

rc

nb

ma

∴ A| =

rnm

cba

∴ If A is a matrix of order (3x2) then A| is a matrix of order (2x3)

When a matrix is multiplied bya scalar K then all its elementshave to be multiplied by K.

Transpose of a matrix is thematrix obtained by changing itsrows into corresponding columnsor columns into correspondingrows.

If the order of a given matrix Ais (m x n) then the order of itstranspose A| is (n x m)

46

Recall (i) : A =

213

132

321

is a symmetric matrix

∴ A| =

213

132

321

(by changing rows into columns)

∴ A| = A or A = A|

Recall (ii) : A =

−−−

043

402

320

is a skew - symmetric matrix

∴ A| =

−−−

043

402

320

Now multiply both sides by (–1), we get

–A| = (–1)

−−−

043

402

320

∴ – A| =

−−−

043

402

320

∴ A = – A|

Recall (iii) :

A = ( )4321 is a row matrix

Now A| =

4

3

2

1

is a column matrix

Think !(A|)| = A

If A = A| then the matrix A is asymmetric Matrix.

If A = –A| then thematrix A is a skewsymmetric matrix.

∴ Transpose of a row matrix isa column matrix

47

Remember : Unit matrix (Identity matrix) is denoted by I

I =

10

01, (2x2) matrix

I =

100

010

001

, (3x3) matrix

Worked Example 1 :

(1) If A =

−−−112

21

x

x is a symmetric matrix. find “x”

Solution : A =

−−−112

21

x

x

∴ A| =

−−

−12

121

x

x

In a symmetric matrix A = A|

∴

−−−112

21

x

x=

−−

−12

121

x

x

∴ x – 2 = 2x – 1

x – 2x = –1 + 2

–x = 1

∴ x = –1

Remember :

(i) If

dc

ba is a symmetric matrix of order (2 x 2) then b = c

(ii) If

oc

bo is a 2 x 2 skew symmetric matrix then b = –c or –b = c

If two matrices are equal thentheir corresponding elements areequal.

48

Worked Example 2 :

If A =

+ 06

20

x

x is a skew symmetric matrix find x

Solution : A =

+ 06

20

x

x is a skew symmetric Matrix

∴ 6 + x = – 2x

x + 2x = –6

3x = –6 ∴ x = –2

3. Addition of Matrices

Consider A =

23

54 and B =

98

76

A and B have the same order (2 x 2) Form a new matrix by adding the correspondingelements of the matrices A and B. The matrix obtained is A + B.

Example : A + B =

23

54 +

98

76

=

++++

9283

7564∴ A + B =

1111

1210

If A and B are two matrices of the same order then the matrix obtainedby adding the corresponding elements of A and B is the sum of the matricesA and B and it is denoted by (A + B).

Example :

A =

nm

qp

yx

and B =

76

54

32

A + B =

nm

qp

yx

+

76

54

32

∴ A + B =

++++++

7n6m

5q4p

3y2x

Remember :To add two matrices, theirorders must be same.

Think :Is the addition ofmatrices commutative?

x

x

x

49

4. Subtraction of matrices

Consider A =

54

73 and B =

73

52

A and B have the same order (2x2). Form a new matrix by subtracting thecorresponding elements of matrix B from the elements of A. The matrix obtained isA – B.

A – B =

54

73 –

73

52

then A – B =

−−−−

7534

5723

∴ A – B =

− 21

21

If A and B are two matrices of the same order then their difference(A–B) is the matrix obtained by subtracting the correspondingelements of matrix B from the elements of matrix A.

Worked Examples :

(1) If A =

41

23 then find A + A|

Solution : A =

41

23

∴ A| =

42

13

A + A| =

41

23 +

42

13

=

++++

4421

1233

∴ A + A| =

83

36

Know this :A – B ≠ B – A

Remember :To subtract two matrices, theirorders must be the same.

Know this :A +A| is a symmetricmatrix.

50

(2) If A =

76

54 then find A – A|

Solution : A =

76

54

∴ A| =

75

64

A – A| =

76

54 –

75

64

=

−−−−

7756

6544

∴ A – A| =

−01

10

(3) If

23

13x +

72

54 =

95

68 then find x

Solution :

23

13x +

72

54 =

95

68

++++

7223

5143x =

95

68

+95

643x =

95

68

∴ 3x + 4 = 8

3x = 8 – 4

3x = 4

x = 3

4

Know this :A – A| is a skewsymmetric matrix.

51

(4) If A =

65

43 and B =

87

21 then find A + 2B.

Solution : B =

87

21

2B = 2

87

21

2B =

1614

42

∴ A + 2B =

65

43+

1614

42

=

++++166145

4423

A + 2B =

2219

85

5. Multiplication of matrices :

Consider A =

765

432 and B =

fe

dc

ba

Multiply the matrix A with matrix B

Then A x B =

765

432

fe

dc

ba

Stage (i) : Multiply the elements of the First Row of A with the corresponding elementsof the First Column of B

a → FR x FC

2 3 4 c = (2 x a) + (3 x c) + (4 x e)

First Row e = 2a + 3c + 4e

52

Stage (ii) :Multiply the elements of First Row of A with the corresponding elementsof the Second Column of B.

b → FR x SC

2 3 4 d = (2 x b) + (3 x d) + (4 x f)

f = 2b + 3d + 4f

Stage (iii): Multiply the elements of second Row of A with the corresponding elementsof the First column of B

a → SR x FC

5 6 7 c = (5 x a) + (6 x c) + (7 x e)

e = 5a + 6c + 7e

Stage (iv) : Multiply the elements of Second Row of A with the corresponding elementsof the Second Column of B

b → SR x SC

5 6 7 d = (5 x b) + (6 x d) + (7 x f)

f = 5b + 6d + 7f

Then AB =

SC.SRFC.SR

SC.FRFC.FR

AB =

++++++++

f7d6b5e7c6a5

f4d3b2e4c3a2

Example 1 : A =

53

42 and B =

98

76

AB =

53

42

98

76

6 = (2 x 6) + (4 x 8)2 4 = 12 + 32

8 = 44

7 = (2 x 7) + (4 x 9)2 4 = 14 + 36

9 = 50

Note : FR = First Row

SC = Second Column

SR = Second Row

FC = First Column

53

6 = (3 x 6) + (5 x 8)3 5 = 18 + 40

8 = 58

7 = (3 x 7) + (5 x 9)3 5 = 21 + 45

9 = 66

∴ AB =

6658

5044(2x2) matrix

Example 2 : A =

43

21 and B =

132

351 then find AB

Solution : AB =

43

21

132

351

=

+++

+++)1x4()3x3()3x4()5x3()2x4()1x3(

)1x2()3x1()3x2()5x1()2x2()1x1(

=

+++

+++

49121583

236541

∴ AB =

132711

5115

Example 3 : If A =

654

321 B =

36

87

Examine if AB exists?

AB =

654

321

36

87

7

1 2 3 6 = (1 x 7) + (2 x 6) + (3 x (?))?

First Row of A has 3 elements. But First Column of B has only 2 elements. Hencemultiplication is not possible. So AB does not exist.

Remember :If the number of columnsof matrix A and thenumber of Rows ofmatrix B are equal thenonly AB exists.

54

CR Rule for multiplication AB

C ≡ R

↓ ↓columns of A Rows of B

6. Properties of matrix multiplication

(1) If A =

41

23 and B =

25

41 then Prove that AB ≠ BA

AB =

41

23

25

41

=

++++

84201

412103

∴ AB =

1221

1613 ..................... (1)

BA =

25

41

41

23

=

++

++810215

16243

∴ BA =

1817

187 .....................(2)

From (1) and (2) we conclude that AB ≠ BA

(2) If A =

43

21 and B =

−54

32 then Prove that (AB)| = B|A|

Solution : AB =

43

21

−54

32

=

−+−+

209166

10382=

−−1122

710

∴ (AB)| =

−− 117

2210 ................ (1)

Note :If A is a matrix of order m x nand B is a matrix of ordern x p then AB exists and itsorder is m x p

55

A| =

42

31 and B| =

−53

42

B|A| =

−53

42

42

31

=

−−++

209103

16682

B|A| =

−− 117

2210 ........ (2)

From (1) and (2) we have (AB)| = B| A|

(iii) If A =

41

23 and I =

10

01 then Prove that AI = IA = A

Solution : AI =

41

23

10

01

=

++++

4001

2003

AI =

41

23 ...... (1)

IA =

10

01

41

23

=

++++

4010

0203

IA =

41

23 ........ (2)

Hence AI = IA = A

Remember :

If A and B are two matricesconformable for multiplicationthen (AB)| = B| A|

Remember :

If A is a square matrix and I isthe unit matrix of the order A thenAI = IA = A.

56

Activity : Taking 3 matrices of order 2x2

Verify (i) A + B = B + A [Commutative property]

(ii) A (BC) = (AB) C [Associative property]

(iii) A(B + C) = AB + AC [Distributive property]

(iv) A(B – C) = AB – AC [Distributive property]

Worked Examples :

(1) Solve for x given that

31

23x +

12

41 =

43

67

Solution :

31

23x +

12

41 =

43

67

++++

1321

4213x =

43

67

+43

613x =

43

67

∴ 3x + 1 = 7 3x = 7 – 1

3x = 6

x =3

6

x = 2

(2) If A =

43

21 then find A2

Solution : A2 = A x A

=

43

21

43

21

=

++++

)4x4()2x3()3x4()1x3(

)4x2()2x1()3x2()1x1(

=

++++166123

8261=

2215

107

57

(3) If A =

01

31 Prove that A2 – A – 3I = 0

Solution : A2 = A x A

=

01

31

01

31

=

++++

)0x0()3x1()1x0()1x1(

)0x3()3x1()1x3()1x1(

=

++++

0301

0331∴ A2 =

31

34

∴ A2 – A – 3I =

31

34 –

01

31 –3

10

01

=

−−−−

0311

3314 –

30

03

=

30

03 –

30

03=

00

00 = 0 ∴ A2 – A – 3I = 0

Exercise : 1.7

(1) If A =

−− 221

312 and B =

−14

23

32

then find 2A + B|

(2) If A =

−24

23 Show that A + A| is a symmetric matrix

(3) If A =

−−

23

14 Show that A – A| is a skew symmetric matrix

(4) If

40

12x +

12

23 =

52

39 find the value of “x”

(5) Solve for x given

54

32x – 2

− 01

1x =

56

10

Note :0 means Zeromatrix

58

(6) If A =

−14

32 and B =

25

23 Show that AB ≠ BA

(7) If A =

43

21 find AA| . (8) If A =

−42

13 find A2

(9) If A =

43

21 and B =

31

12 Verify that (A + B)| = A| + B|

(10)If A =

52

13 Show that A2 – 8A + 13I = 0

(11)If A =

−

620

512 and B =

− 413

201 find AB|

(12)Find x and y given

−25

31

y

x =

−10

32

−1

2x

Sreenivas Ramanujam(Dec 22. 1887 - April 26, 1920)

Ramanujam was one of the most outstanding mathematicianthat India has produced. He worked on the Theory ofNumbers. He stated intuitively many complicated results inmathematics.

Once a great mathematician Prof. Hardy came to India tosee Ramanujam. Prof. Hardy remarked that he has travelledin a taxi with a rather dull number, viz 1729. Ramanujamjumped up and said, oh, No. 1729 is a very interestingnumber. It is the smallest number which can be expressed asthe sum of two cubes in two different ways. viz.

1729 = 13 + 123 = 93 + 103 and the next such number is verylarge.

Sreenivas Ramanujam