M C 1 1 Q l d - 9 172 S e q u e n c e s a n d s e r i e s

Exercise 9A — Recognition of arithmetic sequences 1 a 2, 7, 12, 17, 22.… t2 − t1 = 7 − 2 = 5 t3 − t2 = 12 − 7 = 5 t4 − t3 = 17 − 12 = 5

There is a common difference of 5, therefore d = 5. This is an arithmetic sequence.

b 3, 7, 11, 15, 20.… t2 − t1 = 7 − 3 = 4 t3 − t2 = 11 − 7 = 4 t4 − t3 = 15 − 11 = 4 t5 − t4 = 20 − 15 = 5 There is no common difference,

therefore this is not an arithmetic sequence.

c 0, 100, 200, 300, 400… t2 − t1 = 100 − 0 = 100 t3 − t2 = 200 − 100 = 100 t4 − t3 = 300 − 200 = 100 There is a common difference,

therefore d = 100. This is an arithmetic sequence.

d −123, −23, 77, 177, 277... t2 − t1 = −23 + 123 = 100 t3 − t2 = 77 − −23 = 100 t4 − t3 = 177 − 77= 100 There is a common difference d =

100, so this is an arithmetic sequence.

e 1, 0, −1, −3, 5… t2 − t1 = 0 − 1 = −1 t3 − t2 = −1 − 0 = −1 t4 − t3 = −3 + 1 = −2 This is not an arithmetic

sequence, because there is no common difference.

f 6.2, 9.3, 12.4, 15.5, 16.6… t2 − t1 = 9.3 − 6.2 = 3.1 t3 − t2 = 12.4 − 9.3 = 3.1 t4 − t3 = 15.5 − 12.4 = 3.1 t5 − t4 = 16.6 − 15.5 = 1.1 This is not an arithmetic sequence

because there is no common difference.

g 12

, 1 12

, 2 12

, 3 12

, 4 12

…

t2 − t1 = 1 12

− 12

= 1

t3 − t2 = 2 12

− 1 12

= 1

t4 − t3 = 3 12

− 2 12

= 1

t5 − t4 = 4 12

− 3 12

= 1

Common difference of 1, therefore this is an arithmetic sequence.

h 14

, 34

, 1 14

, 1 34

, 2 14

t2 − t1 = 34

− 14

= 12

t3 − t2 = 1 14

− 34

= 12

t4 − t3 = 1 34

− 1 14

= 12

t5 − t4 = 2 14

− 1 34

= 12

Common difference = 12

Therefore this is an arithmetic sequence.

2 a 7, 12, 17, 22… a = 2, d = 5

c 0, 100, 200, 300, 400 a = 0, d = 100 d −123, −23, 77, 177, 277… a = −123, d = 100

g 1 1 1 1,1 ,2 ,3 . . . . . .2 2 2 2

a = 12

, d = 1

h 1 3 1 3, ,1 , . . . . . .4 4 4 4

a = 14

, d = 12

3 a 2, 7, 12, 17, 22…. a = 2, d = 5 t25 = 2 + (25 − 1)5 t25 = 2 + 24 × 5 t25 = 122 The 25th term is 122

b 0, 100, 200, 300, 400… a = 0, d = 100 t30 = 0 + (30 − 1)100 = 0 + 29 × 100 = 2900 The 30th term is 2900 c 5, −2, −9, −16, −23…. a = 5, d = −7 t33 = 5 + (33 − 1) × −7 = 5 + 32 × −7 = 5 − 224 = −219 The 33rd term is −219

4 a t2 = a + d = 13 (1) t5 = a + 4d = 31 (2) (2) − (1) 3d = 18 d = 6 If d = 6, a = ?

a + 6 = 13 a = 7 t17 = 7 + 16 × 6 = 103 The 17th term is 103

b t2 = a + d = −23 (1) t5 = a + 4d = 277 (2) (2) − (1) 3d = 300 d = 100 If d =100, a = ? a + 100 = −23 a = −123 t20

= −123 + 19 × 100 = 1777 The 20th term is 1777 c t2 = a + d = 0 (1) t6 = a + 5d = −8 (2) (2) − (1) 4d = −8 d = −2 If d = −2, a = ? a − 2 = 0 a = 2 t32 = a + 31 × d = 2 + 31 × −2 = −60 The 32nd term is −60 d t3 = a + 2d = 5 (1) t7 = a + 6d = −19 (2) (2) − (1) 4d = −24 d = −6 If d = −6, a = ? a − 12 = 5 a = 17 t40 = 17 + 39 × −6 = −217 The 40th term is −217 e t9 = a + 3d = 2 (1) t9 = a + 8d = −33 (2) (2) − (1) 5d = −35 d = −7 If d = −7, a = ? a − 21 = 2 a = 23 t26 = 23 + 25 × −7 = −152 The 26th term is −152

5 a 3, 7, 11, 15, 19… a = 3, d = 4

S20 = 202

[2 × 3 + (20 − 1)4]

= 10[6 + 19 × 4] = 820

The sum of the first 20 terms is 820 b 4, 6, 8, 10, 12 a = 4, d = 2

S15 = 152

[2 × 4 + (15 − 1)2]

Chapter 9 — Sequences and series

S e q u e n c e s a n d s e r i e s M C 1 1 Q l d - 9 173

= 7 12

[8 + 14 × 2]

= 270 The sum of the first 15 terms is 270 c −4, −1, 2, 5, 8… a = −4, d = 3

S23 = 232

[2 × −4 + 22 × 3]

= 667 The sum of the first 23 terms is 667

6 −4.3, −2.1, 0.1, 2.3, 4.5 t41 = ? a = −4.3, d = 2.2 t41 = −4.3 + 40 × 2.2 t41 = 83.7 ∴ the answer is A. 7 t2 = a + d = −2 (1) t5 = a + 4d = 2.5 (2) (2) − (1) 3d = 4.5 d = 1.5 ∴ a = −2 − 1.5 a = −3.5 t27 = −3.5 + 26 × 1.5 t27 = 35.5 ∴ the answer is B.

8 0, −3 12

, −7, −10 12

…

a = 0, d = −3 12

S21 = 212

[2 × 0 + 20 × −3 12

]

= −735 ∴ the answer is B. 9 t1 = a = −5.2 t2 = a + d = −6 ∴ d = −6 + 5.2 d = −0.8

S22 = 222

[2 × −5.2 + 21 × −0.8]

= −299.2 ∴ the answer is B. 10 a Yes, an arithmetic sequence where a = 2, d = 2

b No, not an arithmetic sequence c No, not an arithmetic sequence 2, 4, 8…. d Yes, an arithmetic sequence a = ? d = 2 i.e. 37, 39, 41… etc e Using example, Yes, it is an arithmetic sequence 8, 16, 24 a = 8, d = 8

11 a a = 2, d = 2 d a = not specified, d = 2 e a = 8, d = 8

12 a

b

c

d

M C 1 1 Q l d - 9 174 S e q u e n c e s a n d s e r i e s

13 23, 33, 43 a = 23, d = 10 tn = 23 + (n − 1)10 = 23 + 10n − 10 tn = 13 + 10n 14 Row 1 2 3 n 40, 43, 46 … a = 40, d = 3

tn = 40 + (n − 1)3 = 40 + 3n − 3 tn = 37 + 3n 15 12, 15.5, 19, …. a = 12, d = 3.5

a tn = 12 + (n − 1)3.5 = 12 + 3.5n − 3.5 tn = 8.5 + 3.5n b tn = 8.5 + 3.5 × 100 = 358.5 The last post will be 358.5 metres

from the road. 16 t1 = a = 2 S19 = 551

S19 = 192

[4 + 18 × d] = 551

4 + 18d = 58 18d = 54 d = 3

a t19 = 2 + 18 × 3 = 56 The 19th term is 56 b

17 t1 = a = −4 S30 = 2490

S30 = 302

[2 × −4 + 29 × d] = 2490

∴ −8 + 29d = 166 29d = 174 d = 6

a t30 = − 4 + 29 × 6 = 170 The 30th term is 170 b

18 a = 100 d = 75 S15 = ?

S15 = 152

[2 × 100 + 14 × 75]

= $9375 Sam’s total profit is $9375. 19 a = 36 000, d = 1200, S10 = ?

S10 = 102

[36 000 × 2 + 9 × 1200]

= $414 000 George would have earned $414

000. 20 d = 0.8, a = 15, S17 = ?

S17 = 102

[2 × 15 + 16 × 0.8]

= 363.8 cm. The total height of the first 17 steps

is 363.8 cm.

21 a = 250, d = 15, t60 = ? t60 = 250 + 59 × 15 = 1135 After 5 years Paula will have

1135 stamps. 22 1990 1991 1992 2009 3000 3400 3800 …

a = 3000, d = 400, t20 = ? t20 = 3000 + 19 × 400 = $10 600

a The proceeds of the 2009 Fete should be $10 600.

b The total amount would be $136 000.

S20 = 202

(2 × 3000 + 19 × 400)

= $136 000.

Exercise 9B — Geometric sequence 1 a 2, 6, 12, 24, 48…

62

= 3

126

= 2

2412

= 2

4824

= 2

Not a geometric sequence—no common ratio b 1, 4, 16, 64, 256…

2

1

tt

= 41

= 4

3

2

tt

= 164

= 4

4

3

tt

= 6416

= 4

5

4

tt

= 25664

= 4

Geometric sequence r = 4 c −1, −2, −4, −8, −16, …

21

−−

= 2

42

−−

= 2

84

−−

= 2

168

−−

= 2

Geometric sequence r = 2 d 1, 3, −9, 27, −81…

31

= 3

93

− = −3

279−

= −3

8127− = −3

Not a geometric sequence—no common ratio

e 4, −12, 36, −108, 324…

124

− = −3

3612−

= −3

10836

− = −3

324108−

= −3

Geometric sequence r = −3 f −6, 60, −600, 6000, −60 000

606−

= −10

60060

− = −10

6000600−

= −10

60 0006000

− = −10

Geometric sequence r = −10 g −5, 15, −25, 45, −85, …

155−

= −3

12515

− = 53

−

4525−

= 95−

8545

− = 179

−

Not a geometric sequence—no common ratio

h 1.2, 2.4, 4.8, 9.6, 19.2…

2.41.2

= 2

4.82.4

= 2

9.64.8

= 2

19.29.6

= 2

Geometric sequence r = 2 i 7, 3.5, 1.75, 0.875, 0.4375, …

3.57

= 0.5

1.753.5

= 0.5

0.8751.75

= 0.5

0.43750.875

= 0.5

Geometric sequence r = 0.5

S e q u e n c e s a n d s e r i e s M C 1 1 Q l d - 9 175

j 12

, 14

, 18

, 116

, 132

,...

1412

= 12

1814

= 12

11618

= 12

1321

16

= 12

Geometric sequence

r = 12

2 b a = 1, r = 4 c a = 1, r = 2 e a = 4, r = −3 f a = −6, r = −10 h a = 1.2, r = 2

i a = 7, r = 12

j a = 12

, r = 12

3 a 2, 12, 72, 432, 2592…

a = 2, r = 122

= 6

t10 = 2 × 69

= 20 155 392 b 8, 16, 32, 64, 128…

a = 8, r = 168

= 2

t18 = 8 × 217

= 1 048 576 c 5, 15, 45, 135, 405…

a = 5, r = 155

= 3

t11 = 5 × 310

= 295 245 d 2.3, 2.76, 3.312, 3.9744

a = 2.3, r = 2.762.3

= 1.2

t10 = 2.3 × 1.29

= 11.867 494 81 e −2, −8, −32, −128, −512…

a = −2, r = 88

−−

= 4

t9 = −2 × 48

= −131 072 4 a t2 = ar = 6 (1) t5 = ar4 = 162 (2)

(2)(1)

r 3 = 1626

r 3 = 27 r = 3

Sub r = 3 into eq (1) 3a = 6 ∴ a = 2 t10 = 2 × 39

= 39 366 b t2 = ar = 6 (1) t5 = ar4 = 48 (2)

(2)(1)

= r3 = 486

r 3 = 8 r = 2

∴ 2a = 6 a = 3 ∴ t12 = 3 × 211

= 6144 c t2 = ar = 2 (1) t5 = ar4 = 16 (2)

(2)(1)

= r3 = 162

r 3 = 8 r = 2 ∴ 2a = 2 a = 1 ∴ t16 = 1 × 215

= 32 768 d t4 = ar3 = −32 (1) t7 = ar6 = −256 (2)

(2)(1)

= r3 = 25632

−−

r3 = 8 r = 2 ∴ 8a = −32 a = −4 ∴ t14 = −4 × 213

= −32 768 e t4 = ar3 = −192 (1) t7 = ar6 = −12 288 (2)

(2)(1)

= r3 = 12 288192

−−

r3 = 64 r = 4 ∴ 64a = −192 a = −3 ∴ t12 = −3 × 411

= 12 582 912− f t3 = ar2 = 36 (1) t6 = ar5 = −972 (2)

(2)(1)

= r3 = 97236

−

r3 = −27 r = −3 ∴ 9a = 36 a = 4 ∴ t12 = 4 × (−3)11

= −708 588 5 a 1 2 3

5 12.5 31.25 50 000nt t t t

>

a = 5, r = 12.55

= 2.5

tn = 5 × 2.5n − 1

50 000 = 5 × 2.5n − 1

10 000 = 2.5n − 1

log 10 000= (n − 1) log 2.5

log 10 000log 2.5

= n − 1

11.051 = n The 12th term would be the first to

exceed 50 000. b 3.2, 9.6, 28.8, …

a = 3.2, r = 9.63.2

= 3

tn = 3.2 × 3n − 1

1 000 000 = 3.2 × 3n − 1

312 500 = 3n − 1

log312 500log3

= n − 1

12.51667 = n The 13th term would be the first to

exceed 1 000 000. c 5.1, 20.4, 81.6….

a = 5.1, r = 20.45.1

= 4

tn = 5.1 × 4n − 1

100 000 = 5.1 × 4n − 1

19 607.84314 = 4n − 1

log 19 607.84314log 4

= n − 1

8.1295 = n The 9th term would be the first to

exceed 100 000. d 4.3, 9.46, 20.812….

a = 4.3, r = 9.464.3

= 2.2

tn = 4.3 × 2.2n − 1

500, 000 = 4.3 × 2.2n − 1

116 279.0698 = 2.2n − 1

log 116 279.0698log 2.2

= n − 1

15.79312508 = n The 16th term would be the first to

exceed 500 000. 6 a 2, 6, 18, 54, 162

a = 2, r = 62

= 3

S12 = 122(3 1)

3 1−

−

= 531 440 b 5, 35, 245, 1715…

a = 5, r = 355

= 7

S7 = 75(7 1)

7 1−

−

= 686 285 c 1.1, 2.2, 4.4, 8.8…

a = 1.1, r = 2.21.1

= 2

S15= 151.1(2 1)

2 1−

−

= 36 043.7 d 3.1, 9.3, 27.9, 83.7…

a = 3.1, r = 9.33.1

= 3

M C 1 1 Q l d - 9 176 S e q u e n c e s a n d s e r i e s

S11 = 113.1(3 1)

3 1−

−

= 274 576.3 7 a = + ve, r = −2 4, −8, 16, −32, 64, −128

84

− = −2

168−

= −2

3216− = −2

a False, alternate numbers positive b False, 3rd term positive c False, 3rd term > 2nd term d True, 5th term > 6th term, 64 > −128 e False, 4th < 3rd term The answer is D.

8 21, 63, 189, 567…

a = 21, r = 6321

= 3

t12 = 21 × 3n

= 3 720 087 The answer is D. 9 2.25, 4.5, 9, 18, 36…

a = 2.25, r = 4.52.25

= 2

S10 = 102.25(2 1)

2 1−

−

S10 = 2301.75 The answer is B. 10 t2 = ar = −20 (1) t5 = ar4 = −1280 (2)

(2)(1)

⇒ r3= 128020

−−

= 64

r = 4

∴ a = 204

− = −5

∴ S12 = 125(4 1)

4 1− −

−

= −27 962 025 The answer is A. 11 a Day 1 2 3 4 5

1 2 4 8 16

b r = 21

= 42

= 84

= 168

= 2

c 6 7 8 9 10 11 12

16 × 2 16 × 2 × 2 16 × 2 × 2 ×2 16 × 2 × 2 ×2 × 2 16 × 25 16 × 26 16 × 27

2048 people are told the rumour on the 12th day.

12

13 a 1, 2, 4, 8, 16, …n a = 1, r = 2 tn = 1 × 2n − 1

tn = 2n − 1

b t12 = 211

= 2048 There are 2048 cells after 12 divisions.

S e q u e n c e s a n d s e r i e s M C 1 1 Q l d - 9 177

14 10 000, 8500, 7225…

a = 10 000, r = 850010 000

= 0.85

a tn = 10 000 × 0.85n − 1

b t9 = 10 000 × 0.858

= $2724.91 The takings in month 9 will be $2724.91.

15 1, 6, 36… a = 1, r = 6

a tn = 6n − 1

b t5 = 64

By day five, 1296 people should know the rumour. c 4230 = 6n − 1

log 4230log6

= n − 1

5.66 = n After the 6th day all 4230 citizens will know the rumour.

16 60, 64.8, 69.984…

a a = 60, r = 64.860

= 1.08

tn = 60 × 1.08n − 1

b The distance from the water’s edge in year 6 will be t6 = 60 × 1.085

= 88.16 m. c The water will exceed 100 m in the 8th year. 100 = 60 × 1.08n − 1

log 1.66log 1.08

= n − 1

7.6 = n 17 600, 180, 54, 16.2, 4.86 …

a = 600, r = 180600

= 0.3

855 = 600(1 0.3 ).

1 0.3

n−

−

0.9975= 1 − 0.3n

0.3n = 0.0025

n = log 0.0025log 0.3

n = 4.976 Five terms will be needed for the sum to exceed 855.

Exercise 9C — Applications of geometric sequences 1 a

1st year 2nd 3rd 4th … 7th

4 4 × 1.06 4 × (1.06)2 4 × (1.06)6

In his 7th year he should harvest 4 × (1.06)6 = 5.67 tonnes. b 10 tonnes will be reached in 10 = 4 × 1.06n

log 2.5log 1.06

= n

15.72 = n his 17th year.

c S3 = 34(1.06 1)

1.06 1−

−

= 12.7 He can expect to harvest 12.7 tonnes in the first 3 years.

2 a 1st year 2nd year 3rd year … 5th year

1800 1800 × 1.045 1800 × (1.045)2 1800 × (1.045)4

In the 5th the cost will be 1800 × (1.045)4 = $2146.53. b Cost will exceed $2500 2500 = 1800 × 1.045n

log 1.38log 1.045

= n

7.463 = n in the 9th year. c Total costs of keeping her taxi in the first 3 years is:

S3 = 31800(1.045 1)

1.045 1−

−

= $5646.65 3

1st year 2nd year 3rd year

10 000 10 000 × 0.9 10 000 × (0.9)2

⇒ Pn = 10, 000 × (0.9)n − 1

4 a A =13 000 × (1.02)20 R = 84

= 2%

n = 5 × 4 = 20 $13 000 becomes $19 317.32

b Interest earned = 19 317.32 − 13 000 = $6317.32

5 1st year 2nd year 3rd year … 10th year

6000 6000 × 1.1 6000 × (1.1)2 6000 × (1.1)9

a Population in 10th year = 6000 × (1.1)9

= 14 147 b 25 000 = 6000 × (1.1)n

log 4.16log 1.1

= n

14.9 = n In the 16th year the population will reach 25 000.

6 Week 1 Week 2 Week 3

200 200 × 0.85 200 × (0.85)2

a At the end of week 4 Fleas = 200 × (0.85)3

= 122.8 ≈ 123 b 50 = 200 × (0.85)n

0.25 = 0.85n

log 0.25log 0.85

= n

8.5 = n After 10 weeks the fleas should be less than 50.

7 1st year 2nd year 3rd year

2.2 2.2 × 1.09 2.2 × (1.09)2

M C 1 1 Q l d - 9 178 S e q u e n c e s a n d s e r i e s

a In their 4th year they should be: High = 2.2 × (1.09)3

= 2.85 m b 5 = 2.2 × (1.09)n − 1

log 2.2727log 1.09

= n − 1

10.5 = n They should exceed 5 metres after 11 years.

8 Beam 1 Beam 2 Beam 3

0.8 0.8 × 1.03 0.8 × (1.03)2

a The 7th beam will be 0.8 × (1.03)6 = 0.96 m in length. b The first beam to exceed 2 m will be 2 = 0.8 × (1.03)n

log 2.5log 1.03

= n

30.9 = n the 32nd beam.

9 Week 1 Week 2 Week 3

800 800 × 1.02 800 × (1.02)2

1000 = 800 × (1.02)n

log 1.25log 1.02

= n

11.268 = n After 13 weeks The answer is C. 10 1st year 2nd year 3rd year

300 000 300 000 × 1.075 300 000 × (1.075)2

a In the 5th year, they will expect to export: 300 000 × (1.075)4

= $400 640.74 b Exports exceed $ 500 000 in 500 000 = 300 000 × (1.075)n

log 1.66log 1.075

= n

7.063 = n the 9th year. c In the first 7 years total exports equal:

S7 = 7300 000(1.075 1)

1.075 1−

−

= $2 636 196.56 11 1st year 2nd year 3rd year

63 000 63 000 × 0.97 300 000 × (0.97)2

a Attendance in the 5th year = 63 000 × (0.97)4

= 55 773 b Attendance dropped below 50 000 50 000 = 63 000 × (0.97)n

log 0.793650793log 0.97

= n

7.58 = n in the 9th year.

12 a A = 10 000 × (1.1)5

= $16 105.10

b r = 5%, n = 5 × 2 = 10 A = 10 000 × (1.05)10

= $16 288.95

c r = 104

= 2.5%, n = 5 × 4 = 20

A = 10 000 × (1.025)20

= $16 386.16

d r = 1012

, n = 5 × 12 = 60

A = 10 000(1.0083)60

= $16 453.09

13 r = 104

= 2.5%

a A = 20 000 × (1.025)4

= $22 076.26 b n = 3 × 4 = 12 A = 20 000 × (1.075)12

= $26 897.78 c n = 5 × 4 = 20 A = 20 000 × (1.025)20

= $32 772.33 d n = 10 × 4 = 40 A = 20 000 × (1.025)40

= $53 701.28

14 r = 612

= 12

%, n = 3 × 12 = 36

A = 7000 × (1.005)36

= $8376.76

15 r = 84

= 2%, n = ?

7609.45 = 6000 × (1.02)n

log 1.268241667log 1.02

= n

11.99 = n 12 ≈ n

The account was open for 1234

⎛ ⎞⎜ ⎟⎝ ⎠

years.

16 r = 124

= 3%, n = ?

50 670.8 = 40 000 × (1.03)n

log 1.26677log 1.03

= n

8 ≈ n

The account will need to be operating for 2 84

⎛ ⎞⎜ ⎟⎝ ⎠

years.

17 r = 92

= 4.5%, n = ?

15 627.12 = 12 000 × (1.045)n

log 1.3log 1.045

= n

6 ≈ n

The money was in the account for 3 62

⎛ ⎞⎜ ⎟⎝ ⎠

years.

18 r = 1212

= 1%, n = ?

66 277.33 = 60 000(1.01)n

log 1.04622167log 1.01

= n

10 = n The money was invested for 10 months.

S e q u e n c e s a n d s e r i e s M C 1 1 Q l d - 9 179

19 r = 64

= 1.5%

14 500 + 1834.14 = 14 500 × (1.015)n

16 344.14 = 14 500 × (1.015)n

log 1.126492414log 1.015

= n

8 = n

The account was used for 2 84

⎛ ⎞⎜ ⎟⎝ ⎠

years.

Exercise 9D — Finding the sum of an infinite geometric sequence 1 a 50, 25, 12.5, 6.25, 3.125, …

a = 50, r = 2550

= 12

S∞ = 50112

−

S∞ = 100 b 20, 16, 12.8, 10.24, 9.192, …

a = 20, r = 1620

= 45

S∞= 20415

−

= 20 × 51

= 100 c 4, 2.4, 1.44, 0.864, 0.5184, …

a = 4, r = 2.44

= 0.6

S∞= 41 0.6−

= 10

d 1, 13

, 19

, 127

, 181

, …

a = 1, r = 13

S∞ = 1113

−

= 32

e 1, 15

, 125

, 1125

, 1625

, …

a = 1, r = 15

S∞= 1115

−

= 54

f 2, 810

, 32100

, 1281000

, 51210 000

, …

a = 2, r =

8102

= 25

S∞ = 2215

−

S∞ = 235

S∞ = 2 × 53

= 103

= 3 13

g 3, −0.6, 0.12, −0.024, 0.0048, …

a = 3, r = 0.63

− = −0.2

S∞= 31 0.2+

= 2.5 h −12, 7.2, −4.32, 2.592, −1.5552, …

a = −12, r = 7.212−

= −0.6

S∞= 121 0.6

−+

= −7.5 i 120, 48, 19.2, 7.68, 3.072, …

a = 120, r = 48120

= 25

S∞ = 1201 0.4−

= 200 j −50, 5, −0.5, 0.5, −0.05, …

a = −50, r = 550−

= −0.1

S∞ = 501 0.1

−+

= −45 45& &

= −45 511

2 a r = 0.6, S∞ = 25

25 = 1 0.6

a−

∴ a = 10

b r = 0.2, S∞ = 50

50 = 1 0.2

a−

∴ a = 40

c r = 0.25, S∞ = 8

8 = 1 0.25

a−

∴ a = 6

d r = 0.9, S∞ = 120

120 = 1 0.9

a−

M C 1 1 Q l d - 9 180 S e q u e n c e s a n d s e r i e s

∴ a = 12

e r = −0.2, S∞ = 3 13

103

= 1 0.2

a+

∴ 4 = a 4, −0.8, 0.16 f r = −0.5, S∞ = 4

4 = 1 0.5

a+

∴ 6 = a 6, −3, 1.5 g r = −0.8, S∞ = 5

5 = 1 0.8

a+

∴ 9 = a 9, −7.2, 5.76 h r = 0.2, S∞ = −7.5

−7.5 = 1 0.2

a−

∴ a = −6 −6, −1.2, −0.24 i r = 0.6, S∞ = −60

−60 = 1 0.6

a−

∴ a = −24 −24, −14.4, −8.64

j r = −0.3, S∞ = −11 713

15013

− = 1 0.3

a+

∴ −15 = a −15, 4.5, −1.35

3 a a = 12.5, S∞ = 25

25 = 12.51 r−

1 − r = 12.525

1 − r = 0.5 1 − 0.5 = r ∴ 0.5 = r 12.5, 6.25, 3.125

b a = 12.5, S∞ = 50

50 = 12.51 r−

1 − r = 12.550

1 − 0.25 = r 0.75 = r 12.5, 9.375, 7.031 25 c a = 6, S∞ = 8

8 = 61 r−

1 − r = 68

1 − 0.75 = r 0.25 = r 6, 1.5, 0.375

d a = 48, S∞ = 120

120 = 481 r−

1 − r = 48120

1 − 0.4 = r 0.6 = r 48, 28.8, 17.28

e a = 2 49

, S∞ = 3 23

113

=

229

1 r−

1 − r = 22 2

9 3

× 3 1

11 1

1 − r = 23

13

= r

2 49

, 229

× 13

, 2227

× 13

2 49

, 2227

, 2281

4 a 0.5 5& = (0.5 + 0.05 + …)

∴ a = 0.5, r = 0.050.5

= 0.1

S∞ = 0.51 0.1−

0. 5& = 0.50.9

= 59

So 0. 5& = 59

b 0. 8& = 0.8 + 0.08 +

∴ a = 0.8, r = 0.080.8

= 0.1

S∞ = 0.80.9

0. 8& = 0.80.9

= 89

So 0. 8& = 89

c 0. 4& = 0.4 + 0.04 + 0.004 + …

∴ a = 0.4, r = 0.040.4

= 0.1

0. 4& = 0.41 0.1−

= 0.40.9

So 0. 4& = 49

d 1. 3& = 1 + (0.3 + 0.03 + 0.003 + …)

∴ a = 0.3, r = 0.030.3

= 0.1

S e q u e n c e s a n d s e r i e s M C 1 1 Q l d - 9 181

S∞ = 0.31 0.1−

0. 3& = 0.30.9

0. 3& = 39

So 1. 3& = 1 39

= 1 13

e 3. 7& = 3 + (0.7 + 0.07 + 0.007 + …)

∴ a = 0.7, r = 0.070.7

= 0.1

0. 7& = 0.71 0.1−

= 0.70.9

0. 7& = 79

So 3. 7& = 3 79

f 8. 6& = 8 + (0.6 + 0.06 + 0.006 + …)

∴ a = 0.6, r = 0.060.006

= 0.1

0. 6& = 0.61 0.1−

= 0.60.9

0. 6& = 6 29 3⎛ ⎞⎜ ⎟⎝ ⎠

So 8. 6& = 8 23

g 0.14& & = 0.14 + 0.0014 + 0.000014 + …

∴ a = 0.14, r = 0.00140.14

= 0.01

0.14& & = 0.141 0.01−

So 0.14& & = 1499

h 0. 57& & = 0.57 + 0.0057 + 0.000057 + …

∴ a = 0.57, r = 0.00570.57

= .01

0. 57& & = 0.571 .01−

= 5799

i 0.529 = 0.529292929… = 0.5 + 0.029 + 0.00029 + 0.0000029 + …

∴ a = 0.029, r = 0.000290.029

= 0.01

∴ 0.029 = 0.0291 0.01−

= 0.0290.99

= 29990

0.529 = 0.5 + 0.029

= 12

+ 29990

= 495990

+ 29990

= 524990

= 262495

So 0.529 = 262495

j 1.321 = 1 + (0.321 + 0.000321 + …) ∴ a = 0.321,

r = 0.0003210.321

= 0.001

0.321 = 0.3211 0.001−

= 0.3210.999

= 321999

So 1.321 = 1 321999

5 10, 4, 1.6…

r = 410

= 1.64

= 0.4

hence a = 10, r = 0.4

S∞ = 101 0.4−

= 16 23

m

The child will cover 16 23

m. Since her mother is 20m away

she will not reach her. 6 35, 21, 12.6…

r = 2135

= 12.621

= 0.6

hence a = 35, r = 0.6

S∞ = 351 0.6−

= 87.5 m The machine will fall short of spouting by (280 − 87.5 =)

192.5 m. 7 20, 12, 7.2

r = 1220

= 7.21.2

= 0.6

∴ a = 20, r = 0.6

S∞ = 201 0.6−

= 50 mm The nail will be completely hammered in. 8 40 000, 36 800, 33 856…

r = 36 80040 000

= 33 85636 800

= 0.92

∴ a = 40 000, r = 0.92

S∞ = 40 0001 0.92−

= 500 000 She will fall short by $500 000. 9 25, 20, 16…

r = 2025

= 1620

= 0.8

M C 1 1 Q l d - 9 182 S e q u e n c e s a n d s e r i e s

∴ a = 25, r = 0.8

S∞ = 251 0.8−

= 125 mm The system supplies 25 mm too much. 10 a = 1 000 000 S∞ = 12.5 million

a 12 500 000 = 1 000 0001 r−

1 − r = 1 000 00012 500 000

1 − r = 10125

1 − r = 225

1 − 225

= r

2325

= r

b $1 000 000 $920 000 $846 400 $778 688 $716 392.96

c S10 =

10231 000 000 125

23125

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠

−

= $7 070 144.32 After 10 years $7 070 144.32 will be donated in total.

Exercise 9E — Contrasting arithmetic and geometric sequences through graphs 1 Points (1, 0) (2, 1) (3, 2), (4, 3) (5, 4)

1 02 1

−−

= 11

= 1, 2 13 2

−−

= 11

= 1

a constant gradient ⇒ arithmetic sequence b term 1 = a = 0 d = gradient = 1

2 Points (1, 10) (2, 8) (3, 6) (4, 4) (5, 2)

8 10 6 8 4 62 1 3 2 4 32 2 2

m − − −= = =− − −

= − = − = −

a constant gradient ⇒ arithmetic sequence b term 1 = a = 10 gradient = d = −2

3 Terms, 10, 15, 22.5, 33.75, 50.625 a gradient changing ⇒ geometric sequence b term 1 = a = 10

r = 1510

= 22.515

= 33.7522.5

= 50.62533.75

= 1.5

4 Terms 20, 10, 5, 2.5, 1.25 a gradient changing ⇒ geometric sequence b term 1 = a = 20

r = 1020

= 510

= 2.55

= 1.252.5

= 0.5

5 Points (1, 4) 12, 32

⎛ ⎞⎜ ⎟⎝ ⎠

(3, 3) 14, 22

⎛ ⎞⎜ ⎟⎝ ⎠

(5, 2)

a constant gradient ⇒ arithmetic sequence b term 1 = a = 4

d = 3 12

− 4 = 3 − 3 12

= 2 12

− 3 = 2 − 2 12

= − 12

6 100, 80, 64, 51.2, 40.96 a changing gradient ⇒ geometric sequence b term 1 = a = 100

r = 80100

= 6480

= 51.264

= 40.9651.2

= 0.8

7 a Arithmetic, a = 7, d = 2

Term number 1 2 3 4 5 6 7 8

Value of term 7 9 11 13 15 17 19 21

b Geometric, a = 5, r = 12

Term number 1 2 3 4 5 6 7 8

Value of term 5 2.5 1.25 0.625 0.3125 0.15625 0.078125 0.0390625

c Arithmetic, a = −14, d = 3.5

Term number 1 2 3 4 5 6 7 8

Value of term –14 –10.5 –7 –3.5 0 3.5 7 10.5

S e q u e n c e s a n d s e r i e s M C 1 1 Q l d - 9 183

d Arithmetic, a = 32, d = −5

Term number 1 2 3 4 5 6 7 8

Value of term 32 27 22 17 12 7 2 –3

e Geometric, a = 12, r = 103

Term number 1 2 3 4 5 6 7 8

Value of term 12 40 133. 3& 444. 4&

1481.481

4938.3 16 460.9 54 869.68

f Geometric, a = 0.02, r = 6

Term number 1 2 3 4 5 6 7 8

Value of term 0.02 0.12 0.72 4.32 25.92 155.52 933.12 5598.72

8 10, 20, 40, 80, 160

changing gradient a = 10, r = 2010

= 4020

= 8040

= 2

∴ The answer is D. 9 10, −10, 10, −10, 10 Geometric sequence with a =10 and r = −1 ∴ The answer is E. 10

Year 1 2 3

SI 5000 10 1100× ×

1000 1500

A 5500 6000 6500

CI 5000 × 1.1 5000 × (1.1)2 5000 × (1.1)2

A 5500 6050 6655

11

Year 1 2 3

SI 100 000 15100

× 30 000 45 000

15 000

A 115000 130 000 145 000

CI 100 000 × 1.15 132 250 152 087.5

115 000

12 Year 1 2 3

SI 10 000 20 1100

× × 4000 6000

2000

A 12 000 14 000 16 000

CI 10 000 × 1.2 14 400 17 286

12 000

13 un = 10n, vn = 10 × 1.5n − 1

n 1 2 3 4 5

un 10 20 30 40 50

vn 10 15 22.5 33.75 50.625

un is > vn for 3 terms. 14 n 1 2 3 4 5

un 100 80 60 40 20

vn 100 80 64 51.2 40.96

Out of first 5 terms un is never greater.

M C 1 1 Q l d - 9 184 S e q u e n c e s a n d s e r i e s

15 n 1 2 3 4 5

un 120 240 360 480 600

vn 120 180 270 405 607.5

un > vn for 3 terms. 16 n 1 2 3 4 5

un 120 90 60 30 0

vn 100 50 25 12 12

6.25

un > vn for first 4 terms.

Chapter review Multiple choice 1 a

b

c

d

e

The answer is C. 2 −3.6, −2.1 −0.6, 0.9, 2.4… B is an infinite sequence (because …) with a = −3.6 and d = −2.1 + 3.6 = 1.5 The answer is B. 3 a −123, − 23, 77, 177, 277 −23 + 123 = 100 77 + 23 = 100 177 − 77 = 100 277 − 177 = 100 Yes, arithmetic sequence a = −123, d = 100

b −5 14

, −2 14

, 34

, 3 34

, 6 34

94

− + 214

= 3 154

− 34

= 3

34

+ 94

= 3 274

− 154

= 3

Yes, arithmetic sequence a = −5 14

, d = 3

4 −1, 1, 3, 5, 7

111 1 2

1 ( 1)23 1 21 2 25 3 2

2 37 3 2

n

n

at a n dd

nn

t n

= −= + −= + == − + −− == − + −− == −− =

The answer is A.

5 43 7 (42)9 7371 11 2 9

t ad

= − + = −= = − =

The answer is C. 6 t3 = a + 2d = 3.1 (1) t7 = a + 6d = −1.3 (2) (2) − (1) 4d = −1.3 − 3.1 4d = −4.4 d = −1.1 ∴ a = 3.1 − (2 × −1.1) = 5.3 t31 = 5.3 + 30 × −1.1 = −27.7 The answer is B. 7 t2 = a + d = −5 (1) t3 = a + 4d = 16 (2) (2) − (1) 3d = 21 d = 7 ∴ a = −12 226 = −12 + (n − 1)7 226 = −12 + 7n − 7 35 = n 226 is the 35th term. 8

a t15 = 520 + 14 × 40 1080 donations made in 15th year.

b S15 = 152

[2 × 520 + 14 × 40]

= 12 000 12 000 donations over 15 years.

9

a = −16, d = 4

S24 = 242

(2 × −16 + 23 × 4)

= 720 The answer is A. 10 t1 = a = 14 t3 = a + 2d = 8 ∴ 14 + 2d = 8 2d = −6 d = −3

S30 = 302

(2 × 14 + 29 × −3)

= −885 The answer is C. 11 i.e. 3, −6, 12, −24 Let r = −2, then the third term will be a positive number. The answer is B. 12 a 2, −2, 2, −2, 2

22

− = −1 22−

= −1 22

− = −1

22−

= −1

b 2, 4, 6, 8, 10

42

= 2 6 8 102 2 24 6 8

≠ ≠ ≠

c 1, 13

, 19

− , 127

, 181

−

S e q u e n c e s a n d s e r i e s M C 1 1 Q l d - 9 185

131

=

11 9

133

− ≠

11 27

139

− ≠

11 81

1327

−

≠ 13

d 4, −4, 2, −2, 1

42

− = 1 24−

≠ 1 22

− ≠ 1 12−

≠ 1

e 100, 10, 0.1, 0.01, 0.001

10100

= 110

, 0.110

= 1100

, 0.010.1

= 110

, 0.0010.01

= 110

The answer is A.

13 a 5, 52

, 54

, 58

, 516

525

= 52

× 15

= 12

5854

= 58

× 45

= 12

5452

= 54

× 25

= 12

51658

= 516

× 85

= 12

Yes, Geometric, a = 5, r = 12

b −700, −70, −7, 7, 70 Not a Geometric Sequence

because

70700

−−

= 0.1, 770

−−

= 0.1

77−

≠ 0.1, 707

≠ 0.1

14 a = 3.25, r = 6.53.25

= 2

t19 = 3.25 × 218 = 851 968 The answer is B. 15 t3 = ar2 = 19.35 t6 = ar5 = 522.45

r3 = 522.4519.35

= 27

r = 3

∴ a = 19.359

a = 2.15 ∴ t12 = 2.15 × 311 = 380 866.05 The answer is D. 16 2.25, 4.5, 9

a = 2.25, r = 94.5

= 2

1000 = 2.25 × 2n − 1

log 444.4log 2

& = n − 1

9.79 = n

10th term is B. The answer is B. 17 7.2, 8.28, 9.522 7.2, 8.28

7.21.15a r= =

=K

a tn = 7.2 × 1.15n − 1 b t8 = 7.2 × 1.157 = 19.2 tonnes of garbage

collected in 8th year. c 30 = 7.2 × 1.15n − 1

log 4.16log 1.15

& = n − 1

11.2 = n In the 12th year, garbage would

exceed 30 tonnes.

18 8, 4, 2, 1, 12

….

a = 8, r = 48

= 12

S10 =

1018 12112

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠

−

= 15.984375 ≈ 16 The answer is B. 19 t3 = ar2 = 0.9 t6 = ar5 = 7.2

∴ r3 = 7.20.9

= 8

r = 2

a = 0.94

a = 0.225

S12 = 120.225(2 1)

2 1−

−

= 921.375 The answer is C. 20 164, 131.2, 104.96…

a =164, r = 131.2164

= 0.8

800 = 164(1 0.8 )1 0.8

n−−

0.975609756 = 1 − 0.8n 0.8n = 0.024390243

n = log 0.024390243log 0.8

n = 16.642 17 terms are required. 21 1st year 2nd year 6th year

1.2m 1.2 1.05× 51.2 (1.05)×

= 1.53 The answer is A. 22 1 st year 2nd year

60 000 60 000 1.08×

P = 60 000 × 1.08n − 1

100 000 = 60 000 × 1.08n − 1

log 1.66log 1.08

& = n − 1

7.63 = n In the 8th year The answer is C.

23 r = 104

= 2.5%, n = 3 × 4 = 12

a A = 25 000 × (1.025)12 = $33 622.22 b 40 965.41 = 25 000 × (1.025)n

log 1.6386164log 1.025

= n

19.99 = n 20 = n

It would take 5 years 204

⎛ ⎞⎜ ⎟⎝ ⎠

.

24 a = 1, r = 45

S∞ = 1415

− = 1

15

= 5

The answer is E.

25 5 13

= 1 0.5

a+

163

× 1.5 = a

8 = a The answer is D. 26 3. 7& = 3 + (0.7 + 0.07 + …)

∴ a = 0.7, r = 0.070.7

= 0.1

S∞ = 0.71 0.1−

0. 7& = 0.70.9

0. 7& = 79

So 3. 7& = 3 79

27 50, 30, 18…a = 50, r = 3050

= 0.6

S∞ = 501 0.6−

S∞ = 1.25 m The soldier misses by 25 cm. 28 Changing gradient ⇒ Geometric a = 50 50, 75, 112.5, 168.75, 253.125 ⇒ r = 1.5 The answer is D. 29 Constant gradient ⇒ Arithmetic a = 10 10, 5, 0, −5, −10 ⇒ d = −5 The answer is A. 30

n 1 2 3 4

un 10 20 30 40

vn 10 20 40 80

M C 1 1 Q l d - 9 186 S e q u e n c e s a n d s e r i e s

Modelling and problem solving 1

a In the 4th month

15 + 3.5 = 18.5 tonnes of rock will be crushed.

b tn = 8 + (n − 1)3.5 = 8 + 3.5n − 3.5 tn = 4.5 + 3.5n c t60 = 4.5 + 3.5 × 60

= 214.5 tonnes crushed in 60th month.

d 100 = 4.5 + 3.5n 27.28 = n

100 tonnes will be exceeded in the 28th month.

e 3050 = 2n [2 × 8 + (n − 1)3.5]

3050 = 2n [16 + 3.5n − 3.5]

3050 2n

× = 12.5 + 3.5n

6100n

= 12.5 + 3.5n

6100 = 12.5n + 3.5n2 0 = 3.5n2 + 12.5n − 6100 a = 3.5, b = 12.5, c = −6100

n = 212.5 12.5 4 3.5 6100

2 3.5− ± + × ×

×

n = 12.5 85556.257

− ±

n = 12.5 292.57

− + or 12.5 292.52

− −

n = 40 Not feasible After 40 months the environmental

needs to be completed.

2 10, 11, 12.1

a a =10, r = 1.1 1110⎛ ⎞⎜ ⎟⎝ ⎠

b tn = 10 × 1.1n − 1 c t5 = 10 × 1.14

= 14.641 tonnes crushed in 5th month.

d 30 = 10 × 1.1n − 1

log 3log 1.1

= n − 1

12.5 = n 30 tonnes is exceeded in the 13th

month.

e S12 = 1210(1.1 1)

1.1 1−

−

= 213.84 tonnes crushed in 1st year. 3 Using graphical calculator sequences

are:

MTH Quarry 1 Quarry 2

1 8 10

2 11.5 11

3 15 12.1

4 18.5 13.31

5 22 14.641

6 25.5 16.105

7 29 17.716

8 32.5 19.487

9 36 21.436

10 39.5 23.579

11 43 25.937

12 46.5 28.531

13 50 31.384

14 53.5 34.523

15 57 37.975

16 60.5 41.772

17 64 45.95

18 67.5 50.545

19 71 55.599

20 74.5 61.159

21 78 67.273

22 81.5 74.002

23 85 81.403

24 88.5 89.543

The second quarry produces more than the first quarry again in the 24th month.

4 a Consider an investment of $100 at 5% pa simple interest. This represents an arithmetic sequence with

d = 5% of $100 = 5

So the sequence is 100, 105, 110, …..

The value of the investment for each of the first 5 years will be $105, $110, $115, $120, $125.

b Consider the investment of $100 at 5% pa compound interest. This represents a geometric sequence with

r = 1 + 0.05 = 1.05

The sequence is 100 × 1.05, 100 × 1.052, 100 × 1.053, ….

The value of the investment in this case for each of the first 5 years will be $105, $110.25, $115.76, $121.55, $127.63

c Consider a period of ‘n’ years. For simple interest, Amount = 100

+ 5n = ASI For compound interest, Amount =

100 × 1.05n = ACI

Use trial and error to determine a value for n such that ACI ≥ 2ASI

n ASI ACI Ratio ACI : ASI

10 $150 $162.89 1.086 20 $200 $265.33 1.327 30 $250 $432.19 1.729 40 $300 $704.00 2.347 35 $275 $551.60 2.006

So, after 35 years the value of the compound interest investment will be twice that of the simple interest investment.