MC11 Qld-9 172 Sequences and series Exercise 9A — Recognition of arithmetic sequences 1 a 2, 7, 12, 17, 22.… t 2 − t 1 = 7 − 2 = 5 t 3 − t 2 = 12 − 7 = 5 t 4 − t 3 = 17 − 12 = 5 There is a common difference of 5, therefore d = 5. This is an arithmetic sequence. b 3, 7, 11, 15, 20.… t 2 − t 1 = 7 − 3 = 4 t 3 − t 2 = 11 − 7 = 4 t 4 − t 3 = 15 − 11 = 4 t 5 − t 4 = 20 − 15 = 5 There is no common difference, therefore this is not an arithmetic sequence. c 0, 100, 200, 300, 400… t 2 − t 1 = 100 − 0 = 100 t 3 − t 2 = 200 − 100 = 100 t 4 − t 3 = 300 − 200 = 100 There is a common difference, therefore d = 100. This is an arithmetic sequence. d −123, −23, 77, 177, 277... t 2 − t 1 = −23 + 123 = 100 t 3 − t 2 = 77 − −23 = 100 t 4 − t 3 = 177 − 77= 100 There is a common difference d = 100, so this is an arithmetic sequence. e 1, 0, −1, −3, 5… t 2 − t 1 = 0 − 1 = −1 t 3 − t 2 = −1 − 0 = −1 t 4 − t 3 = −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… t 2 − t 1 = 9.3 − 6.2 = 3.1 t 3 − t 2 = 12.4 − 9.3 = 3.1 t 4 − t 3 = 15.5 − 12.4 = 3.1 t 5 − t 4 = 16.6 − 15.5 = 1.1 This is not an arithmetic sequence because there is no common difference. g 1 2 , 1 1 2 , 2 1 2 , 3 1 2 , 4 1 2 … t 2 − t 1 = 1 1 2 − 1 2 = 1 t 3 − t 2 = 2 1 2 − 1 1 2 = 1 t 4 − t 3 = 3 1 2 − 2 1 2 = 1 t 5 − t 4 = 4 1 2 − 3 1 2 = 1 Common difference of 1, therefore this is an arithmetic sequence. h 1 4 , 3 4 , 1 1 4 , 1 3 4 , 2 1 4 t 2 − t 1 = 3 4 − 1 4 = 1 2 t 3 − t 2 = 1 1 4 − 3 4 = 1 2 t 4 − t 3 = 1 3 4 − 1 1 4 = 1 2 t 5 − t 4 = 2 1 4 − 1 3 4 = 1 2 Common difference = 1 2 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 = 1 2 , d = 1 h 13 13 , ,1 , ...... 44 44 a = 1 4 , d = 1 2 3 a 2, 7, 12, 17, 22…. a = 2, d = 5 t 25 = 2 + (25 − 1)5 t 25 = 2 + 24 × 5 t 25 = 122 The 25 th term is 122 b 0, 100, 200, 300, 400… a = 0, d = 100 t 30 = 0 + (30 − 1)100 = 0 + 29 × 100 = 2900 The 30 th term is 2900 c 5, −2, −9, −16, −23…. a = 5, d = −7 t 33 = 5 + (33 − 1) × −7 = 5 + 32 × −7 = 5 − 224 = −219 The 33 rd term is −219 4 a t 2 = a + d = 13 (1) t 5 = a + 4d = 31 (2) (2) − (1) 3d = 18 d = 6 If d = 6, a = ? a + 6 = 13 a = 7 t 17 = 7 + 16 × 6 = 103 The 17 th term is 103 b t 2 = a + d = −23 (1) t 5 = a + 4d = 277 (2) (2) − (1) 3d = 300 d = 100 If d =100, a = ? a + 100 = −23 a = −123 t 20 = −123 + 19 × 100 = 1777 The 20 th term is 1777 c t 2 = a + d = 0 (1) t 6 = a + 5d = −8 (2) (2) − (1) 4d = −8 d = −2 If d = −2, a = ? a − 2 = 0 a = 2 t 32 = a + 31 × d = 2 + 31 × −2 = −60 The 32 nd term is −60 d t 3 = a + 2d = 5 (1) t 7 = a + 6d = −19 (2) (2) − (1) 4d = −24 d = −6 If d = −6, a = ? a − 12 = 5 a = 17 t 40 = 17 + 39 × −6 = −217 The 40 th term is −217 e t 9 = a + 3d = 2 (1) t 9 = a + 8d = −33 (2) (2) − (1) 5d = −35 d = −7 If d = −7, a = ? a − 21 = 2 a = 23 t 26 = 23 + 25 × −7 = −152 The 26 th term is −152 5 a 3, 7, 11, 15, 19… a = 3, d = 4 S 20 = 20 2 [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 S 15 = 15 2 [2 × 4 + (15 − 1)2] Chapter 9 — Sequences and series
15
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Chapter 9 — Sequences and series - Weebly€¦ · MC11 Qld-9 174 Sequences and series 13 23, 33, 43 a = 23, d = 10 tn = 23 + (n − 1)10 = 23 + 10n − 10 tn = 13 + 10n 14 Row 1
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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
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
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)
= $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