p.1 Chp 16 Key Prepared by C.Y. So e.g. 1 p.2 It is given that the first term is 6 and the third term is 2. Let a = first term and d = common difference. Thus, 6 a and 2 2 d a . It follows that 2 2 ) 6 ( d 4 8 2 6 2 2 2 2 ) 6 ( d d d d (a) Now, the general term d n a n T ) 1 ( ) ( n n n 4 10 4 4 6 ) 4 )( 1 ( ) 6 ( (b) Using (a), ) 15 ( 4 10 ) 15 ( T 50 60 10 ) 15 ( 4 10 e.g. 2 p.3 It is given that the 17 5 T and 77 20 T . Let a = first term and d = common difference. Thus, ) 2 ( ...... 17 4 ) 1 ( ...... 77 19 d a d a . (1)(2), 60 15 d 4 d Sub 4 d into (2), 17 ) 4 ( 4 a 1 a The 1st term is 1. The 2nd term = 1 + 4 = 5. The 3rd term = 5 + 4 = 9. The 4th term = 9+4 = 13.
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p.1
Chp 16 Key Prepared by C.Y. So
e.g. 1 p.2
It is given that the first term is 6 and the third term is 2.
Let a = first term and d = common difference.
Thus, 6a and 22 da .
It follows that 22)6( d
4
82
622
22)6(
d
d
d
d
(a) Now, the general term dnanT )1()(
n
n
n
410
446
)4)(1()6(
(b) Using (a), )15(410)15( T
50
6010
)15(410
e.g. 2 p.3
It is given that the 175 T and 7720 T .
Let a = first term and d = common difference.
Thus,
)2( ...... 174
)1( ...... 7719
da
da.
(1)(2), 6015 d
4d
Sub 4d into (2), 17)4(4 a
1a
The 1st term is 1. The 2nd term = 1 + 4 = 5. The 3rd term = 5 + 4 = 9. The 4th term = 9+4 = 13.
p.2
15 23
d d d d
e.g. 3 p.3
It is given that the 501 T , 472 T and 443 T .
Let a = first term and d = common difference.
Thus, 50a and 5047 d
3
(a) Let 17)( nT . (b) Let 0)( nT .
12
336
31753
173350
17)3)(1()50(
17)1(
n
n
n
n
n
dna
3
217
3
53
353
03350
0)3)(1()50(
0)1(
n
n
n
n
n
dna
Thus, the 12th term is 17. Thus, the 17th term is positive and the 18th term is negative.
2
4850
)3(16)50(17
T
Thus, the smallest positive term is 2.
(Actually, there is a much quicker way. Think of it yourself.)
e.g. 4 p.4
The difference between 15 and 23
= 23 (15)
= 38
Let d = the common difference between two terms.
5.9
384
d
d
The first term added in The second term added in The third term added in
5.5
5.915
4
5.95.5
5.13
5.94
p.3
20 40
d d d d d d
5 27
d d d
x y
e.g. 5 p.5
The difference between 20 and 40
= 40 20
= 20
Let d = the common difference between two terms.
3
13
3
10
206
d
d
d
The first term added in The second term added in The third term added in
3
123
3
1320
3
226
3
13
3
123
303
13
3
226
The fourth term added in The fifth term added in
3
133
3
1330
3
236
3
13
3
133
e.g. 6 p.5
The difference between 27 and 5
= 27 5
= 22
Let d = the common difference between two terms.
3
17
3
22
223
d
d
d
3
112
3
175
x
3
219
3
17
3
112
y
p.4
e.g. 7 p.6
By formula: Recall that with a = first term, d = common difference and n = number of terms
Sum to n terms ])1(2[2
)( dnan
nS
For ...131074 ,
424
538
)3158(8
)]3)(116()4(2[2
)16()16(
S
Alternative: 49...131074
424
161363
162
13123
16)12...4321(3
16)48...12963(
e.g. 8 p.6
Let the three numbers be dy , y and dy .
As the sum of them equals 18, As the product of them equals 210,
6
183
18)()(
y
y
dyydy
1
1
3536
35)6)(6(
210)6)(6)(6(
2
2
d
d
d
dd
dd
Therefore, the three numbers are 5, 6 and 7.
e.g. 9 p.6
For the G.S., let a = the first term and r = the common ratio.
As the sixth term is 1, 15 ar .
As the third term is 8, 82 ar .
p.5
(a) Thus, we have 8
12
5
ar
ar
2
1
8
1)(
8
1
3
1
3
13
3
r
r
r
The common ratio is 2
1 .
(b) Sub 2
1r into 82 ar .
32
84
1
82
12
a
a
a
The first term is 32.
(c) The tenth term
16
12
1)1(
2
132)1(
2
1)32(
4
9
9
9
ar
e.g. 10 p.8
For the G.S., let a = the first term and r = the common ratio.
As the seventh term is 48, 486 ar .
As the fourth term is 6, 63 ar .
Thus, we have 6
483
6
ar
ar
p.6
2
8)(
8
3
13
13
3
r
r
r
Sub 2r into 486 ar .
4
364
48
4864
482
48)2(6
6
a
a
a
a
a
Thus, the first term is 4
3 and the common ratio is 2.
e.g. 11 p.8
For the G.S., let a = the first term and r = the common ratio.
It is given that the sum of the 1st term and the 3rd term is 40.
402 ara … (1)
It is given that the 2nd term is larger than the 1st term by 8.
8 aar … (2)
From (1), 402 ara
40)1( 2 ra
From (2), 8 aar
8)1( ra
Thus, we have 8
40
)1(
)1( 2
ra
ra.
5)1(
)1( 2
r
r
065
551
)1(51
2
2
2
rr
rr
rr
2r or 3r .
When 2r , 8a . In this case, the first term is 8 and the common ratio is 2.
When 3r , 4a . In this case, the first term is 4 and the common ratio is 3.
p.7
e.g. 12 p.9
For the G.S., let a = the first term and r = the common ratio.
After adding in five terms between 2 and 1458,
2 is still the 1st term and 1458 becomes the 7th term.
Thus, we have 14586 ar … (1)
and 2a … (2).
(1) (2), we have
3
729
2
1458
6
6
r
r
a
ar
The first number added in The second number added in The third number added in
6
32
18
36
54
318
The fourth number added in The fifth number added in
162
354
486
3162
e.g. 13 p.9
For the G.S., let a = the first term and r = the common ratio.
It is given that x, 6, 18, y are in a G.S.
3
186
r
r
2
63
x
x
54
54
318
y
y
y
p.8
e.g. 14 p.10
By formula: Recall that with a = first term, r = common difference and n = number of terms
Sum to n terms )1(1
)( nrr
anS
For ...1263 , 2r .
3069
)11024(3
)12(12
3
])2(1[)2(1
)3()10(
10
10
S
Alternative: 923...1263
3069
)11024(312
123
21
213
)21(
)21)(2...221(3
)2...221(3
23...232313
10
10
92
92
92
e.g. 15 p.10
For the G.S., 15, 30, 60, … , 3840,
let a = the first term and r = the common ratio.
(a) Thus, 15a .
2
3015
r
r
Let 38401 nar .
9
81
22
2562
15
3840
15
2)15(
38402)15(
81
1
1
1
n
n
n
n
n
n
The number of terms of the G.S. is 9.
p.9
(b) )3840(...)60()30()15(
7665
51115
)12(12
)15(
)21()2(1
)15(
9
9
e.g. 16 p.11
By formula: Recall that with a = first term and r = common difference
Sum to infinity r
aS
1
For ...139 , 39 r
3
1r
2
272
393
2 9
3
11
)9(
S
e.g. 17 p.11
By formula: Recall that with a = first term and r = common difference
Sum to infinity r
aS
1
4
4164
1
16
4
31
16
a
a
a
a
p.10
e.g. 18 p.12
(a)
6
1 36
i
i
29
7362
76
3
136
)654321(3
166
3
)6(6
3
)5(6
3
)4(6
3
)3(6
3
)2(6
3
)1(6
(b)
n
i
inS
1 36)(
36
12482
98
3
148
)8...321(3
186
3
16
36
36)8(
8
1
8
1
8
1
8
1
8
1
ii
ii
i
i
i
iS
Ex16.4 1. p.13
Let a = first term and d = common difference.
As the 9th term is 25, As the 5th term is 13,
258 da … (1) 134 da … (2)
(1) (2), 13254 d
124 d
3d
13)3(4 a , 1a .
The 60th term = (1) + (59)(3)
= 178
p.11
Ex16.4 2. p.13
(a) For the A.S., 1002, 1005, 1008, … , 1998
can be rewritten as (1000+2), (1000+5), (1000+8), … , (1000+998)
can be rewritten as (999+3), (999+6), (999+9), … , (999+999)
can be rewritten as (999+31), (999+32), (999+33), … , (999+3333)
Therefore, there are 333 terms in the A.S. 1002, 1005, 1008, … , 1998 which are divisible by 3.
(Alternative: You may let 1002a , 3d , 1998)1()(T dnan and solve for n)
(b) For the A.S., 1000, 1004, 1008, … , 2000
can be rewritten as 1000, (1000+41), (1000+42), … , (1000++4250)
Therefore, there are 251 terms in the A.S. 1000, 1004, 1008, … , 2000 which are divisible by 4.
(c) For the A.S., 1008, 1020, 1032, … , 1992
can be rewritten as (996+121), (996+122), (996+123), … , (996+1283)
Therefore, there are 83 terms in the A.S. 1008, 1020, 1032, … , 1992 which are divisible by 12.
Thus, the number of integers between 1000 and 2000 (including 1000 and 2000)
501
83251333
Ex16.4 3. p.13
(In solving this problem, the following technique has been employed : 2231 TTTT )
As a is the arithmetic mean between 8 and b, As 12 is the arithmetic mean between a and b,
it means 8, a, b forms an A.S.. it means a, 12, b forms an A.S..
aab 8 (Using the fact that 2231 TTTT )
1212 ba
82 ba … (1) 24 ba … (2)
(1) + (2), 2483 a
3
32a
3
210a
Sub 3
210a into (2), 24
3
210
b
3
113b
p.12
Ex16.4 4. p.13
For the G.S., let a = the first term and r = the common ratio.
As the sum of the second and third term is 20, As the sum of the fourth and fifth term is 320,
202 arar 32043 arar
20)1( rar ... (1) 320)1(3 rar … (2)
(2) (1), 20
320
)1(
)1(3
rar
rar
4
162
r
r
Sub 4r into (1), 20)5)(4( a
1a
The general term T(n)
1
1
1
4
)4)(1(
n
n
nar
Ex16.4 5. p.14
Let the three numbers be a, da and da 3 .
(a) As they form a G.S., ))(()3)(( dadadaa (Using the fact that 2231 TTTT )
2
222 23
dad
dadaada
02 dad 0)( dad Thus, 0d or da .
As the sum of the three numbers is 14,
1443
14)3()(
da
dadaa
When 0d ,
3
14
143
a
a
Thus, the three numbers are 3
14,
3
14,
3
14.
The common ratio is 1.
p.13
When ad ,
2
147
a
a
Thus, the three numbers are 2, 4, 8.
The common ratio is 2.
(b) The common difference of the A.S. is 0 or 2.
(c) When 0d , the three numbers are 3
14,
3
14,
3
14.
When 2d , the three numbers are 2, 4, 8.
Ex16.4 6. p.14
(In solving this problem, the following technique has been employed :
2231 TTTT for an A.S.
2231 TTTT for a G.S.)
As 1, x, y are in a G.S., As x, y, 15 are in an A.S.,
2
1
xy
xxy
152
15
yx
yyx
Combining the above two equations,
1520
152
15)(2
2
2
2
xx
xx
xx
2
5x or 3x
When 2
5x , When 3x ,
2
2
5
y
9
)3( 2
y
4
25
Ex16.4 7. p.14
(This question is cancelled)
p.14
Ex16.4 8. p.14
By observation: 2, 22, 204, 2006, 20008, 200010, …