Chapter 1 worked solutions – Sequences and series

Solutions to Exercise 1D
1a 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑚 − 5 = 17 − 𝑚
2𝑚 = 22
𝑚 = 11

1b 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑚 − 32 = 14 − 𝑚
2𝑚 = 46
𝑚 = 23

1c 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑚 − (−12) = (−50) − 𝑚
2𝑚 = −62
𝑚 = −31

1d 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑚 − (−23) = 7 − 𝑚
2𝑚 = −16
𝑚 = −8

1e 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
22 − 𝑚 = 32 − 22
22 − 𝑚 = 10
𝑚 = 12

1f 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
−5 − (−20) = 𝑚 − (−5)
15 = 𝑚 + 5
𝑚 = 10

𝑇2 𝑇
2a = 𝑇3
𝑇1 2
𝑔 18
=
2 𝑔
𝑔2 = 36
𝑔 = 6 or − 6

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Chapter 1 worked solutions – Sequences and series

𝑇2 𝑇
2b = 𝑇3
𝑇1 2
𝑔 3
=𝑔
48
2
𝑔 = 144
𝑔 = 12 or − 12

𝑇2 𝑇
2c = 𝑇3
𝑇1 2
𝑔 −90
=
−10 𝑔
𝑔2 = 900
𝑔 = 30 or − 30

𝑇2 𝑇
2d = 𝑇3
𝑇1 2
𝑔 −2
=
−98 𝑔
2
𝑔 = 196
𝑔 = 14 or − 14

𝑇2 𝑇
2e = 𝑇3
𝑇1 2
20 80
= 20
𝑔
20
=4
𝑔
4𝑔 = 20
𝑔=5

𝑇2 𝑇
2f = 𝑇3
𝑇1 2
4 𝑔
=
−1 4
𝑔 = −16

3a i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − 4 = 16 − 𝑥
2𝑥 = 20
𝑥 = 10

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Chapter 1 worked solutions – Sequences and series

𝑇2 𝑇
3a ii = 𝑇3
𝑇1 2
𝑥 16
=
4 𝑥
2
𝑥 = 64
𝑥 = 8 or − 8

3b i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − 1 = 49 − 𝑥
2𝑥 = 50
𝑥 = 25

𝑇2 𝑇
3b ii = 𝑇3
𝑇1 2
𝑥 49
=
1 𝑥
2
𝑥 = 49
𝑥 = 7 or − 7

3c i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − 16 = 25 − 𝑥
2𝑥 = 41
1
𝑥 = 20 2

𝑇2 𝑇
3c ii = 𝑇3
𝑇1 2
𝑥 25
=
16 𝑥
2
𝑥 = 400
𝑥 = 20 or − 20

3d i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − (−5) = −20 − 𝑥
2𝑥 = −25
1
𝑥 = −12 2

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Chapter 1 worked solutions – Sequences and series

𝑇2 𝑇
3d ii = 𝑇3
𝑇1 2
𝑥 −20
=
−5 𝑥
2
𝑥 = 100
𝑥 = 10 or − 10

3e i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
10 − 𝑥 = 50 − 10
10 − 𝑥 = 40
𝑥 = −30

𝑇2 𝑇
3e ii = 𝑇3
𝑇1 2
10 50
= 10
𝑥
10
=5
𝑥
5𝑥 = 10
𝑥=2

3f i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
12 − 𝑥 = 24 − 12
12 − 𝑥 = 12
𝑥=0

𝑇2 𝑇
3f ii = 𝑇3
𝑇1 2
12 24
= 12
𝑥
12
=2
𝑥
2𝑥 = 12
𝑥=6

3g i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
−1 − 𝑥 = 1 − (−1)
−1 − 𝑥 = 2
𝑥 = −3

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Chapter 1 worked solutions – Sequences and series

𝑇2 𝑇
3g ii = 𝑇3
𝑇1 2
−1 1
= −1
𝑥
1
− 𝑥 = −1
𝑥=1

3h i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
6 − 𝑥 = −12 − 6
6 − 𝑥 = −18
𝑥 = 24

𝑇2 𝑇
3h ii = 𝑇3
𝑇1 2
6 −12
=
𝑥 6
6
= −2
𝑥
−2𝑥 = 6
𝑥 = −3

3i i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
30 − 20 = 𝑥 − 30
10 = 𝑥 − 30
𝑥 = 40

𝑇2 𝑇
3i ii = 𝑇3
𝑇1 2
30 𝑥
= 30
20
900
𝑥= 20
𝑥 = 45

3j i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
24 − (−36) = 𝑥 − 24
60 = 𝑥 − 24
𝑥 = 84

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Chapter 1 worked solutions – Sequences and series

𝑇2 𝑇
3j ii = 𝑇3
𝑇1 2
24 𝑥
− 36 = 24
576
𝑥=− 36
𝑥 = −16

3k i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
1
−3 − (− 4) = 𝑥 − (−3)
3
−2 4 = 𝑥 + 3
3
𝑥 = −5 4

𝑇2 𝑇
3k ii = 𝑇3
𝑇1 2
−3 𝑥
1 = −3
−
4
𝑥
− 3 = 12
𝑥 = −36

3l i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
−7 − 7 = 𝑥 − (−7)
−14 = 𝑥 + 7
𝑥 = −21

𝑇2 𝑇
3l ii = 𝑇3
𝑇1 2
−7 𝑥
= −7
7
𝑥
− 7 = −1
𝑥=7

4a 𝑇𝑛 = 7 + (𝑛 − 1)𝑑
Put 𝑇6 = 42
7 + (6 − 1)𝑑 = 42
7 + 5𝑑 = 42
5𝑑 = 42 − 7
5𝑑 = 35
𝑑=7

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Chapter 1 worked solutions – Sequences and series

𝑇𝑛 = 7 + 7(𝑛 − 1)
𝑇𝑛 = 7𝑛
𝑇1 = 7
𝑇2 = 14
𝑇3 = 21
𝑇4 = 28
𝑇5 = 35
𝑇6 = 42

4b 𝑇𝑛 = 27𝑟 𝑛−1
Put 𝑇4 = 8
8 = 27𝑟 4−1
8 = 27𝑟 3
8
𝑟3 =
27
2
𝑟=
3
𝑇1 = 27
𝑇2 = 18
𝑇3 = 12
𝑇4 = 8

4c 𝑇𝑛 = 48 + (𝑛 − 1)𝑑
Put 𝑇5 = 3
3 = 48 + (5 − 1)𝑑
−45 = 4𝑑
1
𝑑 = −11
4
1
𝑇𝑛 = 48 − 11 (𝑛 − 1)
4
𝑇1 = 48
3
𝑇2 = 36
4
1
𝑇3 = 25
2
1
𝑇4 = 14
4

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Chapter 1 worked solutions – Sequences and series

𝑇5 = 3

4d 𝑇𝑛 = 48𝑟 𝑛−1
Put 𝑇5 = 3
3 = 48𝑟 5−1
1
= 𝑟4
16
1
𝑟=±
2
1
When 𝑟 = 2

𝑇𝑛 = 3(2)𝑛−1
𝑇1 = 48
𝑇2 = 24
𝑇3 = 12
𝑇4 = 6
𝑇5 = 3
1
When 𝑟 = − 2

𝑇𝑛 = 3(−2)𝑛−1
𝑇1 = 48
𝑇2 = −24
𝑇3 = 12
𝑇4 = −6
𝑇5 = 3

5a 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑
𝑇10 = 18 gives 18 = 𝑎 + 9𝑑 (1)
𝑇20 = 48 gives 48 = 𝑎 + 19𝑑 (2)
Subtract (1) from (2):
30 = 10𝑑
𝑑=3
Substitute 𝑑 = 3 into (1):
18 = 𝑎 + 9 × 3
𝑎 = −9

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Chapter 1 worked solutions – Sequences and series

5b 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑
𝑇5 = 24 gives 24 = 𝑎 + 4𝑑 (1)
𝑇9 = −12 gives −12 = 𝑎 + 8𝑑 (2)
Subtract (1) from (2):
−36 = 4𝑑
𝑑 = −9
Substitute 𝑑 = −9 into (1):
24 = 𝑎 + 4 × −9
𝑎 = 60

5c 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑
𝑇4 = 6 gives 6 = 𝑎 + 3𝑑 (1)
𝑇12 = 34 gives 34 = 𝑎 + 11𝑑 (2)
Subtract (1) from (2):
28 = 8𝑑
1
𝑑 = 32
1
Substitute 𝑑 = 3 2 into (1):
1
6 = 𝑎 + 3 × 32
1
𝑎 = −4 2

𝑇6 𝑎𝑟 6−1 𝑎𝑟 5
6a = 𝑎𝑟 3−1 = 𝑎𝑟 2 = 𝑟 3
𝑇3

𝑇6 128
= =8
𝑇3 16
𝑟3 = 8
𝑟=2
𝑇3 = 𝑎𝑟 3−1 = 𝑎𝑟 2 = 𝑎(2)2 = 4𝑎 and 𝑇3 = 16
4𝑎 = 16
𝑎=4

𝑇6 𝑎𝑟 6−1 𝑎𝑟 5
6b = 𝑎𝑟 2−1 = = 𝑟4
𝑇2 𝑎𝑟

𝑇6 27
= = 81
𝑇2 1
3
𝑟 4 = 81
𝑟 = 3 or − 3

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Chapter 1 worked solutions – Sequences and series

When 𝑟 = 3,
𝑇6 = 𝑎𝑟 6−1 = 𝑎𝑟 5 = 𝑎(3)5 = 243𝑎 and 𝑇6 = 27
243𝑎 = 27
1
𝑎=9

When 𝑟 = −3,
𝑇6 = 𝑎𝑟 6−1 = 𝑎𝑟 5 = 𝑎(−3)5 = −243𝑎 and 𝑇6 = 27
−243𝑎 = 27
1
𝑎 = −9

𝑇9 𝑎𝑟 9−1 𝑎𝑟 8
6c = 𝑎𝑟 5−1 = 𝑎𝑟 4 = 𝑟 4
𝑇5

𝑇9 24
= =4
𝑇5 6
𝑟4 = 4

𝑟 = √2 or −√2

When 𝑟 = √2,
4
𝑇5 = 𝑎𝑟 5−1 = 𝑎𝑟 4 = 𝑎(√2) = 4𝑎 and 𝑇5 = 6

4𝑎 = 6
3
𝑎=2

When 𝑟 = −√2,
4
𝑇5 = 𝑎𝑟 5−1 = 𝑎𝑟 4 = 𝑎(−√2) = 4𝑎 and 𝑇5 = 6

4𝑎 = 6
3
𝑎=2

7a 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑
𝑇3 = 7 gives 7 = 𝑎 + 2𝑑 (1)
𝑇7 = 31 gives 31 = 𝑎 + 6𝑑 (2)
Subtract (1) from (2):
24 = 4𝑑
𝑑=6

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Chapter 1 worked solutions – Sequences and series

Substitute 𝑑 = 6 into (1):

7=𝑎+2×6
𝑎 = −5
𝑇8 = −5 + 7 × 6 = 37

7b 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑
𝑑 = −7
𝑇10 = 3 gives 3 = 𝑎 + 9 × −7
3 = 𝑎 − 63
𝑎 = 66
𝑇2 = 66 − 7 = 59

7c 𝑇𝑛 = 𝑎𝑟 𝑛−1
𝑟=2
𝑇6 = 6 gives 6 = 𝑎 × 25
6 3
𝑎 = 32 = 16
3 3
𝑇2 = 16 × 21 = 8

8a 3𝑛 > 1 000 000

ln 3𝑛 > ln 1 000 000
𝑛 ln 3 > ln 1 000 000
ln 1 000 000
𝑛>
ln 3
𝑛 > 12.575 …
The smallest integer solution is 𝑛 = 13.

8b 5𝑛 < 1 000 000

ln 5𝑛 < ln 1 000 000
𝑛 ln 5 < ln 1 000 000
ln 1 000 000
𝑛<
ln 5
𝑛 < 8.584 …
The largest integer solution is 𝑛 = 8.

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Chapter 1 worked solutions – Sequences and series

8c 7𝑛 > 1 000 000 000

ln 7𝑛 > ln 1 000 000 000
𝑛 ln 7 > ln 1 000 000 000
ln 1 000 000 000
𝑛>
ln 7
𝑛 > 10.64 …
The smallest integer solution is 𝑛 = 11.

8d 12𝑛 < 1 000 000 000

ln 12𝑛 < ln 1 000 000 000
𝑛 ln 12 < ln 1 000 000 000
ln 1 000 000 000
𝑛<
ln 12
𝑛 < 8.339 …
The largest integer solution is 𝑛 = 8.

𝑇2 4 𝑇3 8
9a = 2 = 2 and =4=2
𝑇1 𝑇2
Hence the sequence is a GP with 𝑟 = 2 and 𝑎 = 2.
𝑇𝑛 = 𝑎𝑟 𝑛−1
𝑇𝑛 = 2 × 2𝑛−1
= 21 × 2𝑛−1
= 2𝑛

9b 𝑇𝑛 < 1 000 000

2𝑛 < 1 000 000
log10 2𝑛 < log10 1 000 000
𝑛 log10 2 < log10 1 000 000
log10 1 000 000
𝑛< log10 2
𝑛 < 19.93 …
Hence there are 19 terms less than 1 000 000.

9c 𝑇𝑛 < 1 000 000 000

2𝑛 < 1 000 000 000
log10 2𝑛 < log10 1 000 000 000
𝑛 log10 2 < log10 1 000 000 000
log10 1 000 000 000
𝑛< log10 2

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Chapter 1 worked solutions – Sequences and series

𝑛 < 29.8973 …
Hence there are 29 terms less than 1 000 000 000.

9d 𝑇𝑛 < 1020
2𝑛 < 1020
log10 2𝑛 < log10 1020
𝑛 log10 2 < 20
20
𝑛 < log 2
10
𝑛 < 66.43 …
Hence there are 66 terms less than 1020 .

9e Using the answers to parts b and c, there are 10 terms between 1 000 000 and
1 000 000 000.

9f Using the answers to parts c and d, there are 37 terms between 1 000 000 000
and 1020 .

𝑇2 14 1
10a = 98 = 7
𝑇1

1 1 𝑛−1
This is a GP with 𝑎 = 98, 𝑟 = 7 so 𝑇𝑛 = 98 (7) .

𝑇𝑛 > 10−6
1 𝑛−1
98 ( ) > 10−6
7
1 𝑛−1 10−6
( ) >
7 98
1 𝑛−1 10−6
ln ( ) > ln
7 98
1 10−6
(𝑛 − 1) ln > ln
7 98
10−6
ln 1
98
𝑛−1< 1 (Note that ln 7 < 0, hence we must reverse the sign)
ln
7

𝑛 − 1 < 9.46 …
𝑛 < 10.46 …

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Chapter 1 worked solutions – Sequences and series

Hence there are 10 terms greater than 10−6.

𝑇2 5 1
10b = 25 = 5
𝑇1

1 1 𝑛−1
This is a GP with 𝑎 = 25, 𝑟 = 5 so 𝑇𝑛 = 25 (5) .

𝑇𝑛 > 10−6

1 𝑛−1
25 ( ) > 10−6
5
1 𝑛−1 10−6
( ) >
5 25
1 𝑛−1 10−6
ln ( ) > ln
5 25
1 10−6
(𝑛 − 1) ln > ln
5 25
10−6
ln
𝑛−1< 25
1
ln
5
𝑛 − 1 < 10.58 …
𝑛 < 11.58 …
Hence there are 11 terms greater than 10−6.

𝑇2 0.9
10c = = 0.9
𝑇1 1

This is a GP with 𝑎 = 1, 𝑟 = 0.9 so 𝑇𝑛 = (0.9)𝑛−1 .

𝑇𝑛 > 10−6
(0.9)𝑛−1 > 10−6
(𝑛 − 1) ln 0.9 > ln 10−6
ln 10−6
𝑛−1<
ln 0.9
𝑛 − 1 < 131.13 …
𝑛 < 132.13 …
Hence there are 132 terms greater than 10−6 .

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Chapter 1 worked solutions – Sequences and series

11a This is a GP with 𝑇𝑛 = 0.97𝑛 , where 𝑇𝑛 is the intensity of the light, and 𝑛
represents the number of sheets of glass.
For 50 sheets of glass:
𝑇50 = 0.9750 = 0.22 or 22%
Hence the light’s intensity is reduced by 1 − 22% = 78% after passing through
50 sheets of glass.

11b 𝑇𝑛 < 0.01

0.97𝑛 < 0.01
ln(0.97𝑛 ) < ln(0.01)
𝑛 ln(0.97) < ln(0.01)
ln(0.01)
𝑛 > ln(0.97)
𝑛 > 151.19 …
𝑇152 = 0.97152 = 0.975 … %
Hence a minimum of 152 sheets of glass are required to reduce the light’s
intensity to below 1%.

12a 𝑇6 + 𝑇8 = 44
𝑎 + (6 − 1)𝑑 + 𝑎 + (8 − 1)𝑑 = 44
2𝑎 + 12𝑑 = 44 (1)
𝑇10 + 𝑇13 = 35
𝑎 + (10 − 1)𝑑 + 𝑎 + (13 − 1)𝑑 = 35
𝑎 + 9𝑑 + 𝑎 + 12𝑑 = 35
2𝑎 + 21𝑑 = 35 (2)
9𝑑 = −9 (2)−(1)
𝑑 = −1 (3)
2𝑎 + 12(−1) = 44 (3) in (1)
2𝑎 = 56
𝑎 = 28
So 𝑎 = 28 and 𝑑 = −1.

12b 𝑇2 + 𝑇3 = 4
𝑎𝑟 2−1 + 𝑎𝑟 3−1 = 4
𝑎𝑟 1 + 𝑎𝑟 2 = 4
𝑎𝑟(1 + 𝑟) = 4 (1)

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Chapter 1 worked solutions – Sequences and series

𝑇4 + 𝑇5 = 36
𝑎𝑟 3 + 𝑎𝑟 4 = 36
𝑎𝑟 3 (1 + 𝑟) = 36 (2)
𝑟2 = 9 (2) ÷ (1)
𝑟 = ±3
When 𝑟 = −3, (3)
𝑎(−3)(1 − 3) = 4 (3) in (1)
6𝑎 = 4
2
𝑎=
3
When 𝑟 = 3 (4)
𝑎(3)(1 + 3) = 4 (4) in (1)
12𝑎 = 4
1
𝑎=
3
2 1
So 𝑎 = 3 and 𝑟 = −3, or 𝑎 = 3 and 𝑟 = 3

12c 𝑇4 + 𝑇6 + 𝑇8 = −6
As this is an AP, 𝑇8 = 𝑇6 + 2𝑑 and 𝑇4 = 𝑇6 − 2𝑑, hence
𝑇6 + 2𝑑 + 𝑇6 + 𝑇6 − 2𝑑 = −6
3𝑇6 = −6
𝑇6 = −2

13a 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
17 − (𝑥 − 1) = (𝑥 + 15) − 17
18 − 𝑥 = 𝑥 − 2
2𝑥 = 20
𝑥 = 10
The numbers are: 9, 17, 25.

13b 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
(𝑥 − 4) − (2𝑥 + 2) = 5𝑥 − (𝑥 − 4)
−𝑥 − 6 = 4𝑥 + 4
5𝑥 = −10

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Chapter 1 worked solutions – Sequences and series

𝑥 = −2
The numbers are: −2, −6, −10.

13c 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
5 − (𝑥 − 3) = (2𝑥 + 7) − 5
8 − 𝑥 = 2𝑥 + 2
3𝑥 = 6
𝑥=2
The numbers are: −1, 5, 11.

13d 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − (3𝑥 − 2) = (𝑥 + 10) − 𝑥
−2𝑥 + 2 = 10
−2𝑥 = 8
𝑥 = −4
The numbers are: −14, −4, 6.

𝑇3 𝑇
14a = 𝑇2
𝑇2 1

𝑥 𝑥+1
=
𝑥+1 𝑥
𝑥 2 = (𝑥 + 1)2
𝑥 2 = 𝑥 2 + 2𝑥 + 1
2𝑥 = −1
1
𝑥=−
2
1 1 1
The numbers are: − 2 , 2 , − 2.

𝑇3 𝑇
14b = 𝑇2
𝑇2 1

5−𝑥 2
=
2 2−𝑥
(5 − 𝑥)(2 − 𝑥) = 4
10 − 7𝑥 + 𝑥 2 = 4
𝑥 2 − 7𝑥 + 6 = 0
(𝑥 − 1)(𝑥 − 6) = 0
𝑥 = 1 or 𝑥 = 6
When 𝑥 = 1, the numbers are: 1, 2, 4.
When 𝑥 = 6, the numbers are: −4, 2, −1.

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Chapter 1 worked solutions – Sequences and series

15a i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
24 − 𝑥 = 96 − 24
24 − 𝑥 = 72
𝑥 = −48
The numbers are: −48, 24, 96.

𝑇2 𝑇
15a ii = 𝑇3
𝑇1 2
24 96
= 24
𝑥
24
=4
𝑥
4𝑥 = 24
𝑥=6
The numbers are: 6, 24, 96.

15b i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − 0.2 = 0.000 02 − 𝑥
2𝑥 = 0.200 02
𝑥 = 0.100 01
The numbers are: 0.2, 0.100 01, 0.000 02.

𝑇2 𝑇
15b ii = 𝑇3
𝑇1 2
𝑥 0.000 02
=
0.2 𝑥
2
𝑥 = 0.000 004
𝑥 = 0.002 or − 0.002
The numbers are: 0.2, 0.002, 0.000 02 or 0.2, −0.002, 0.000 02.

15c i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
0.2 − 𝑥 = 0.002 − 0.2
−𝑥 = −0.398
𝑥 = 0.398
The numbers are: 0.398, 0.2, 0.002.

𝑇2 𝑇
15c ii = 𝑇3
𝑇1 2
0.2 0.002
=
𝑥 0.2
0.2
𝑥
= 0.01
0.01𝑥 = 0.2

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Chapter 1 worked solutions – Sequences and series

𝑥 = 20
The numbers are: 20, 0.2, 0.002.

15d i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
(𝑥 + 1) − (𝑥 − 4) = (𝑥 + 11) − (𝑥 + 1)
5 = 10 FALSE
Hence, these numbers cannot form an AP.

𝑇2 𝑇
15d ii = 𝑇3
𝑇1 2
𝑥+1 𝑥+11
=
𝑥−4 𝑥+1
(𝑥 + 1)(𝑥 + 1) = (𝑥 + 11)(𝑥 − 4)
𝑥 2 + 2𝑥 + 1 = 𝑥 2 + 7𝑥 − 44
2𝑥 + 1 = 7𝑥 − 44
5𝑥 = 45
𝑥=9
The numbers are: 5, 10, 20.

15e i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
(𝑥 + 2) − (𝑥 − 2) = (5𝑥 − 2) − (𝑥 + 2)
4 = 4𝑥 − 4
4𝑥 = 8
𝑥=2
The numbers are: 0, 4, 8.

𝑇2 𝑇3
15e ii =
𝑇1 𝑇2
𝑥+2 5𝑥−2
=
𝑥−2 𝑥+2
(𝑥 + 2)(𝑥 + 2) = (5𝑥 − 2)(𝑥 − 2)
𝑥 2 + 4𝑥 + 4 = 5𝑥 2 − 12𝑥 + 4
0 = 4𝑥 2 − 16𝑥
0 = 𝑥 2 − 4𝑥
0 = 𝑥(𝑥 − 4)
𝑥 = 0 or 𝑥 = 4
The numbers are: −2, 2, −2 or 2, 6, 18.

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Chapter 1 worked solutions – Sequences and series

15f i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − (√5 + 1) = (√5 − 1) − 𝑥
2𝑥 = 2√5
𝑥 = √5
The numbers are: √5 + 1, √5, √5 − 1.

𝑇2 𝑇
15f ii = 𝑇3
𝑇1 2
𝑥 √5−1
=
√5+1 𝑥
2
𝑥 = (√5 + 1)(√5 − 1)
𝑥2 = 5 − 1 = 4
𝑥 = −2 or 2
The numbers are: √5 + 1, −2, √5 − 1 or √5 + 1, 2, √5 − 1.

15g i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − √2 = √8 − 𝑥
𝑥 − √2 = 2√2 − 𝑥
2𝑥 = 3√2
3
𝑥 = 2 √2
3
The numbers are: √2, 2 √2, √8.

𝑇2 𝑇
15g ii = 𝑇3
𝑇1 2
𝑥 √8
=
√2 𝑥
2
𝑥 = √2 × 8
𝑥 2 = √16
𝑥2 = 4
𝑥 = −2 or 2
The numbers are: √2, −2, √8 or √2, 2, √8.

15h i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − 24 = 26 − 𝑥
2𝑥 = 24 + 26
2𝑥 = 80
𝑥 = 40
The numbers are: 24 , 40, 26 .

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Chapter 1 worked solutions – Sequences and series

𝑇2 𝑇
15h ii = 𝑇3
𝑇1 2
𝑥 26
=
24 𝑥
𝑥 2 = 210
𝑥 = 25 or −25
The numbers are: 24 , 25 , 26 or 24 , −25 , 26 .

15i i 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
𝑥 − 7 = −7 − 𝑥
𝑥=0
The numbers are: 7, 0, −7.

𝑇2 𝑇
15i ii = 𝑇3
𝑇1 2
𝑥 −7
=
7 𝑥
𝑥 2 = −49 This is a false statement.
These numbers cannot form a GP.

16a For a GP:

𝑇3 𝑇2
=
𝑇2 𝑇1
1 𝑏
=
𝑏 𝑎
𝑎 = 𝑏2 (1)
For an AP:
𝑇3 − 𝑇2 = 𝑇2 − 𝑇1
10 − 𝑎 = 𝑎 − 𝑏
2𝑎 − 𝑏 − 10 = 0 (2)
2𝑏 2 − 𝑏 − 10 = 0 (1) in (2)
(2𝑏 − 5)(𝑏 + 2) = 0
5 1 5 2 1
Hence 𝑏 = 2 = 2 2 and 𝑎 = (2) = 6 4 or 𝑏 = −2 and 𝑎 = (−2)2 = 4.

16b For a GP:

𝑇3 𝑇2
=
𝑇2 𝑇1
𝑎+𝑏 1
=
1 𝑎
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Chapter 1 worked solutions – Sequences and series

𝑎2 + 𝑎𝑏 = 1 (1)
For an AP:
𝑇3 − 𝑇2 = 𝑇2 − 𝑇1
1 1
𝑎−𝑏− = −𝑏
2 2
𝑎=1 (2)
1+𝑏 =1 (2) in (1)
𝑏=0
Hence 𝑎 = 1, 𝑏 = 0

17a For an AP, the terms must be of the form

𝑇1 = 𝑥 − 𝑑
𝑇2 = 𝑥
𝑇2 = 𝑥 + 𝑑
as given.
For a GP:
𝑇2 𝑇
= 𝑇3
𝑇1 2

𝑥 𝑥+𝑑
=
𝑥−𝑑 𝑥
𝑥 2 = (𝑥 − 𝑑)(𝑥 + 𝑑)
𝑥 2 = 𝑥 2 − 𝑑2
0 = −𝑑 2
𝑑2 = 0
𝑑=0
Hence 𝑇1 = 𝑇2 = 𝑇3 = 𝑥 so all terms are the same.

17b In an AP:
𝑇1 = 𝑎
𝑇4 = 𝑎 + 3𝑑
𝑇7 = 𝑎 + 6𝑑
𝑇7 − 𝑇4 = 𝑎 + 6𝑑 − (𝑎 + 3𝑑) = 3𝑑
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Chapter 1 worked solutions – Sequences and series

𝑇4 − 𝑇1 = 𝑎 + 3𝑑 − (𝑎) = 3𝑑
So 𝑇7 − 𝑇4 = 𝑇4 − 𝑇1and thus 𝑇1 , 𝑇4 and 𝑇7 form an AP as they have the same
common difference.

17c In an GP:
𝑇1 = 𝑎
𝑇4 = 𝑎𝑟 3
𝑇7 = 𝑎𝑟 6
𝑇7 𝑎𝑟 6 𝑟 6
= = = 𝑟3
𝑇4 𝑎𝑟 3 𝑟 3
𝑇4 𝑎𝑟 3 𝑟 3
= = = 𝑟3
𝑇1 𝑎 1
𝑇 𝑇
So 𝑇7 = 𝑇4 and thus 𝑇1 , 𝑇4 and 𝑇7 form a GP as they have the same common ratio.
4 1

18a For an AP, the terms must be of the form:

𝑇1 = 𝑎
𝑇2 = 𝑎 + 𝑑
𝑇4 = 𝑎 + 3𝑑
To form a GP, terms must have the same common ratio so
𝑎 + 3𝑑 𝑎 + 𝑑
=
𝑎+𝑑 𝑎
𝑎(𝑎 + 3𝑑) = (𝑎 + 𝑑)2
𝑎2 + 3𝑎𝑑 = 𝑎2 + 2𝑎𝑑 + 𝑑 2
𝑎𝑑 − 𝑑 2 = 0
𝑑(𝑎 − 𝑑) = 0
Hence 𝑑 = 0 or 𝑎 = 𝑑.
If 𝑑 = 0, then the AP sequence is constant.
If 𝑎 = 𝑑 then 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑎 = 𝑛𝑎 = 𝑛𝑇1 and so the terms are positive integer
multiples of the first term.

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Chapter 1 worked solutions – Sequences and series

18b For an AP, the terms must be of the form:

𝑇1 = 𝑎
𝑇2 = 𝑎 + 𝑑
𝑇5 = 𝑎 + 4𝑑
To form a GP, terms must have the same common ratio so
𝑎 + 4𝑑 𝑎 + 𝑑
=
𝑎+𝑑 𝑎
𝑎(𝑎 + 4𝑑) = (𝑎 + 𝑑)2
𝑎2 + 4𝑎𝑑 = 𝑎2 + 2𝑎𝑑 + 𝑑 2
2𝑎𝑑 − 𝑑2 = 0
𝑑(2𝑎 − 𝑑) = 0
Hence 𝑑 = 0 or 𝑑 = 2𝑎.
If 𝑑 = 0, then the AP sequence is constant.
If 𝑑 = 2𝑎 then 𝑇𝑛 = 𝑎 + (𝑛 − 1) × 2𝑎 = (2𝑛 − 1)𝑎 = (2𝑛 − 1)𝑇1 and so the terms
are odd positive integer multiples of the first term.

18c For an AP, the terms must be of the form:

𝑇1 = 𝑎
𝑇2 = 𝑎𝑟
𝑇4 = 𝑎𝑟 3
To form an AP, terms must have the same common difference so
𝑎𝑟 3 − 𝑎𝑟 = 𝑎𝑟 − 𝑎
𝑟3 − 𝑟 = 𝑟 − 1
𝑟 3 − 2𝑟 + 1 = 0
(𝑟 − 1)(𝑟 2 − 𝑟 − 1) = 0
Hence 𝑟 = 1 or 𝑟 2 − 𝑟 − 1 = 0
Using the quadratic formula:

1 ± √12 − 4(1)(−1)
𝑟=
2

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Chapter 1 worked solutions – Sequences and series

1 ± √5
=
2
1 1 1 1
So 𝑟 = 1, + 2 √5 or − 2 √5
2 2

18d For the GP:

𝑇1 = 𝑎, 𝑇2 = 𝑎𝑟, 𝑇3 = 𝑎𝑟 2
𝑆1 = 𝑎, 𝑆2 = 𝑎 + 𝑎𝑟, 𝑆3 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2
However, each term is one more than the sum of all the previous terms.
𝑇2 = 𝑎𝑟 and 𝑇2 = 1 + 𝑆1 = 1 + 𝑎 so 𝑎𝑟 = 1 + 𝑎
1+𝑎
or 𝑟 = (1)
𝑎

𝑇3 = 𝑎𝑟 2 and 𝑇3 = 1 + 𝑆2 = 1 + 𝑎 + 𝑎𝑟 so
𝑎𝑟 2 = 1 + 𝑎 + 𝑎𝑟 (2)
Substituting (1) into (2):

1+𝑎 2 1+𝑎
𝑎×( ) =1+𝑎+𝑎×
𝑎 𝑎
(1 + 𝑎)2
=1+𝑎+1+𝑎
𝑎
(1 + 𝑎)2 = 𝑎(2 + 2𝑎)
1 + 2𝑎 + 𝑎2 = 2𝑎 + 2𝑎2
1 = 𝑎2
𝑎 = ±1
Substituting in (1):
1+𝑎 1+1
When 𝑎 = 1, 𝑟 = = =2
𝑎 1
1+𝑎 1−1
When 𝑎 = −1, 𝑟 = = = 0 (Not possible values for the GP.)
𝑎 −1

So the GP is 1, 2, 4, 8, …

19a The square of any real number cannot be negative so (𝑎 − 𝑏)2 ≥ 0.

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Chapter 1 worked solutions – Sequences and series

19b (𝑎 − 𝑏)2 ≥ 0
𝑎2 − 2𝑎𝑏 + 𝑏 2 ≥ 0
𝑎2 − 2𝑎𝑏 + 𝑏 2 + 4𝑎𝑏 ≥ 0 + 4𝑎𝑏
𝑎2 + 2𝑎𝑏 + 𝑏 2 ≥ 4𝑎𝑏
(𝑎 + 𝑏)2 ≥ 4𝑎𝑏

𝑎 + 𝑏 ≥ 2√𝑎𝑏
1
(𝑎 + 𝑏) ≥ 2𝑎𝑏
2
𝑚≥𝑔
Hence 𝑔 ≤ 𝑚

1
19c Set 𝑎 = 1 for both sets of sequences so 𝑚 = 2 (1 + 𝑏) or 𝑏 = 2𝑚 − 1
1
If 𝑚 = 1 then 𝑔 = 2 (1 + 1) = 1 and 𝑏 = 2 × 1 − 1 = 1.

Hence we have that 1, 1, 1 is trivially an AP and a GP.

1
If 𝑚 = 5 then 𝑔 = 2 (1 + 5) = 3 and 𝑏 = 2 × 5 − 1 = 9.

Hence we have that 1, 5, 9 is an AP whilst 1, 3, 9 is a GP.

1
20a Put 𝑇13 = 2 𝑇1

1
𝑎𝑟 12 = 𝑎
2
1
𝑟 12 =
2
1
1 12
𝑟= (2) (taking 𝑟 > 0 as pipes do not have negative lengths)

1 7 7
7 1 12 1 12 2
20b 𝑇8 = 𝑎𝑟 = 𝑇1 ((2) ) = 𝑇1 (2) ≑ 0.667𝑇1 ≑ 3 𝑇1

4
4 1 12 4
20c 𝑇5 = 𝑎𝑟 = 𝑇1 (2) ≑ 0.7937𝑇1 ≑ 5 𝑇1

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Chapter 1 worked solutions – Sequences and series

3
20d Put 𝑇𝑛 = 4 𝑇1

3
𝑎𝑟 𝑛−1 = 𝑎
4
1 𝑛−1
1 12 3
(( ) ) =
2 4
1−𝑛 3
2 12 =
4
By trial and error, the closest integer solution is 𝑛 = 6 so the sixth pipe is about
three-quarters of the length of the first pipe.
5
Put 𝑇𝑛 = 6 𝑇1

5
𝑎𝑟 𝑛−1 = 𝑎
6
1 𝑛−1
1 12 5
(( ) ) =
2 6
1−𝑛 5
2 12 =
6

By trial and error, the closest integer solution is 𝑛 = 4 so the fourth pipe is about
five-sixths of the length of the first pipe.
2
2 1 12 8
20e 𝑇3 = 𝑎𝑟 = 𝑇1 (2) ≑ 0.8908𝑇1 ≑ 9 𝑇1
1
1 12 17
𝑇2 = 𝑎𝑟 1 = 𝑇1 ( ) ≑ 0.9439𝑇1 ≑ 𝑇
2 18 1

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Chapter 1 worked solutions – Sequences and series

21a-d

21e ∠𝑂𝑇𝑀 = 90° because it is an angle in a semi-circle, so 𝑂𝑇 is a tangent.

As 𝑂𝑇 is a tangent, it follows that ∠𝑂𝑇𝐴 = ∠𝑂𝐵𝑇 as angle of a triangle inscribed

in a circle is equal to the angle formed by the opposing chord and tangent.
∠𝐵𝑂𝑇 = ∠𝐴𝑂𝑇 as they are the same angle. Hence 𝑂𝐴𝑇 and 𝑂𝑇𝐵 are equiangular
and thus similar. As 𝑂𝐴𝑇 and 𝑂𝑇𝐵 are similar triangles, by the equal ratio of
𝑂𝑇 𝑂𝐴
sides in similar triangles 𝑂𝐵 = 𝑂𝑇 . Thus 𝑂𝑇 2 = 𝑂𝐴 × 𝑂𝐵.

© Cambridge University Press 2019 28