14 Arithmetic and Geometric Sequences and their Summation
14 Arithmetic and Geometric Sequences and their Summation
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14 Arithmetic and Geometric Sequences and their Summation
Activity
Activity 14.1 (p. 161)
1. (a) 2, 2, 2, 2
(b) 3, 3, 3, 3
(c) 5, 5, 5, 5
(d) –5, –5, –5, –5
2. They are equal.
Activity 14.2 (p. 184)
1. (a) S(10) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
(b) S(10) = 55
2. No
Activity 14.3 (p. 198)
1. / Range of R / R / n / RnR < –1 / –2
(or any other reasonable answers) / 8 / 256
(or any other reasonable answers)
15 / –32 768
(or any other reasonable answers)
very large / very large or very small
–1 < R < 1 / (or any other reasonable answers) / 8 / (or any other reasonable answers)
15 / (or any other reasonable answers)
very large / very close to zero
R > 1 / 2
(or any other reasonable answers) / 8 / 256 (or any other reasonable answers)
15 / 32 768 (or any other reasonable answers)
very large / very large
2. (a) very large or very small
(b) very close to zero
(c) very large
Follow-up Exercise
p. 158
1. (a) T(1) = 5(1) =
T(2) = 5(2) =
T(3) = 5(3) =
T(4) = 5(4) =
(b) T(1) = 12 – 1 =
T(2) = 22 – 1 =
T(3) = 32 – 1 =
T(4) = 42 – 1=
(c) T(1) =
T(2) =
T(3) =
T(4) =
2. (a) (i) 21, 26
(ii) ∵ T(1) = 1 = 5(1) – 4
T(2) = 6 = 5(2) – 4
T(3) = 11 = 5(3) – 4
T(4) = 16 = 5(4) – 4
∴ T(n) =
(iii) ∵ T(n) = 5n – 4
∴ T(10) = 5(10) – 4 = 46
and T(15) = 5(15) – 4 = 71
∴ The 10th term and the 15th term of the sequence are 46 and 71 respectively.
(b) (i) 81, 243
(ii) ∵ T(1) = 1 = 31 – 1
T(2) = 3 = 32 – 1
T(3) = 9 = 33 – 1
T(4) = 27 = 34 – 1
∴ T(n) =
(iii) ∵ T(n) = 3n – 1
∴ T(10) = 310 – 1 = 39
and T(15) = 315 – 1 = 314
∴ The 10th term and the 15th term of the sequence are 39 and 314 respectively.
(c) (i)
(ii) ∵ T(1) =
T(2) =
T(3) =
T(4) =
∴ T(n) =
(iii) ∵ T(n) =
∴ T(10) =
and T(15) =
∴ The 10th term and the 15th term of the sequence are and respectively.
p. 165
1. (a) T(2) – T(1) = 3 – 1 = 2
T(3) – T(2) = 5 – 3 = 2
T(4) – T(3) = 7 – 5 = 2
∴ It is an arithmetic sequence with common difference 2.
(b) T(2) – T(1) = –10 – (–13) = 3
T(3) – T(2) = –7 – (–10) = 3
T(4) – T(3) = –4 – (–7) = 3
∴ It is an arithmetic sequence with common difference 3.
(c) T(2) – T(1) = 4 – 2 = 2
T(3) – T(2) = 8 – 4 = 4
∴ It is not an arithmetic sequence.
(d) T(2) – T(1) = log 4 – log 2
= 2 log 2 – log 2
= log 2
T(3) – T(2) = log 8 – log 4
= 3 log 2 – 2 log 2
= log 2
T(4) – T(3) = log 16 – log 8
= 4 log 2 – 3 log 2
= log 2
∴ It is an arithmetic sequence with common difference log 2.
2. (a) Let a and d be the first term and the common difference respectively.
∵ a = 2 and d = 5 – 2 = 3
∴ T(n) = a + (n – 1)d
= 2 + (n – 1)(3)
= 3n – 1
∴ T(12) = 3(12) – 1
=
(b) Let a and d be the first term and the common difference respectively.
∵ a = 10 and d = 14 – 10 = 4
∴ T(n) = a + (n – 1)d
= 10 + (n – 1)(4)
= 6 + 4n
∴ T(12) = 6 + 4(12)
=
(c) Let a and d be the first term and the common difference respectively.
∵ a = 6 and d = 4 – 6 = –2
∴ T(n) = a + (n – 1)d
= 6 + (n – 1)(–2)
= 8 – 2n
∴ T(12) = 8 – 2(12)
=
(d) Let a and d be the first term and the common difference respectively.
∵ a = –25 and d = –22 – (–25) = 3
∴ T(n) = a + (n – 1)d
= –25 + (n – 1)(3)
= 3n – 28
∴ T(12) = 3(12) – 28
=
3. (a) Let a and d be the first term and the common difference respectively.
T(9) = a + 8d = 22 ……(1)
T(13) = a + 12d = 34 ……(2)
(2) – (1), 4d = 12
d = 3
By substituting d = 3 into (1), we have
a + 8(3) = 22
a = –2
∴ The first term and the common difference are –2 and 3 respectively.
(b) T(n) = a + (n – 1)d
= –2 + (n – 1)(3)
=
(c) ∵ T(k) = 73
∴ –5 + 3k = 73
k =
4. (a) T(n) = a + (n – 1)d
= –24 + (n – 1)(5)
=
(b) T(7) = – 29 + 5(7)
=
T(12) = – 29 + 5(12)
=
(c) ∵ The mth term is the first positive term of the sequence.
∴ T(m) > 0
i.e. – 29 + 5m > 0
m >
∵ m is the number of terms, it must be an integer.
∴ m =
p. 170
1. (a) Arithmetic mean =
= –6
(b) Arithmetic mean =
= 82
2. (a) Let d1 be the common difference of the arithmetic sequence to be formed.
The arithmetic sequence formed is:
5, 5 + d1, 5 + 2d1, 17
∵ The 4th term is also given by 5 + 3d1.
∴ 5 + 3d1 = 17
d1 = 4
∴ The two required arithmetic means are 9 and 13.
(b) Let d2 be the common difference of the arithmetic sequence to be formed.
The arithmetic sequence formed is:
5, 5 + d2, 5 + 2d2, 5 + 3d2, 17
∵ The 5th term is also given by 5 + 4d2.
∴ 5 + 4d2 = 17
d2 = 3
∴ The three required arithmetic means are 8, 11 and 14.
(c) Let d3 be the common difference of the arithmetic sequence to be formed.
The arithmetic sequence formed is:
5, 5 + d3, 5 + 2d3, 5 + 3d3, 5 + 4d3, 5 + 5d3, 17
∵ The 7th term is also given by 5 + 6d2.
∴ 5 + 6d3 = 17
d3 = 2
∴ The five required arithmetic means are 7, 9, 11, 13 and 15.
3. ∵ x is the arithmetic mean between 6 and y,
and 18 is the arithmetic mean between y and 22.
From (2), we have
36 = y + 22
y =
By substituting y = 14 into (1), we have
x =
= 10
p. 176
1. (a) = 2
∴ It is not a geometric sequence.
(b)
∴ It is a geometric sequence with common ratio.
(c)
∴ It is not a geometric sequence.
(d) = 2
∴ It is not a geometric sequence.
2. (a) Let a and R be the first term and the common ratio respectively.
∵ a = 1 and R == 2
∴ T(n) = aRn – 1
= 1(2)n – 1
= 2n – 1
∴ T(6) = 26 – 1
=
(b) Let a and R be the first term and the common ratio respectively.
∵ a = 3 and R == 3
∴ T(n) = aRn – 1
= 3(3)n – 1
= 3n
∴ T(6) = 36
=
(c) Let a and R be the first term and the common ratio respectively.
∵ a = –3 and R == –2
∴ T(n) = aRn – 1
= –3(–2)n – 1
=(–2)n
∴ T(6) = (–2)6
=
(d) Let a and R be the first term and the common ratio respectively.
∵ a = 128 and R =
∴ T(n) = aRn – 1
= 128
= –256
∴ T(6) = –256
=
3. (a) Let a and R be the first term and the common ratio respectively.
∵ a = –8, R = and T(n) = aRn – 1
∴ T(n) =
(b) T(4) = –24 – 4
=
T(6) = –24 – 6
=
4. (a) Let a and R be the first term and the common ratio respectively.
T(3) = aR2 = 1 ………(1)
T(8) = aR7 = 243 ………(2)
(2) ¸ (1), R5 = 243
R = 3
By substituting R = 3 into (1), we have
a(3)2 = 1
a =
∴ The first term and the common ratio are and 3 respectively.
(b) T(n) = aRn – 1
=× 3n – 1
=
(c) T(7) + T(9) = 37 – 3 + 39 – 3
=
p. 181
1. (a) Geometric mean =
= 15
(b) Geometric mean =
= 42
2. (a) Let R be the common ratio of the geometric sequence to be formed.
The geometric sequence formed is:
∵ The 4th term is also given by R3.
∴
R =
∴ The two required geometric means are and .
(b) Let r be the common ratio of the geometric sequence to be formed.
The geometric sequence formed is:
∵ The 5th term is also given by r4.
∴
r =
∴ The three required geometric means are , and or , and .
3. ∵ x, x + 3, x + 9 are in geometric sequence.
∴ x + 3 is the geometric mean between x and x + 9.
∴ (x + 3)2 = x(x + 9)
x2 + 6x + 9 = x2 + 9x
3x = 9
x =
p. 183
1. (a) 1 + 3 + 5 + 7 + 9
(b) 25
2. (a) 2 + 4 + 8 + 16 + 32 + 64
(b) 126
3. (a) 1 + 4 + 9 + 16 + 25 + 36 + 49
(b) 140
p. 188
1. (a) ∵ a = 1, d = 5 – 1 = 4 and n = 10
∴ S(10) =[2(1) + (10 – 1)(4)]
=
(b) ∵ a = 3, d = 8 – 3 = 5 and n = 12
∴ S(12) =[2(3) + (12 – 1)(5)]
=
(c) ∵ a = 28, d = 26 – 28 = –2 and n = 15
∴ S(15) =[2(28) + (15 – 1)( –2)]
=
2. (a) ∵ a = –5, l = 9 and n = 8
∴ S(8) =
=
(b) ∵ a = 100, d = –4 and n = 12
∴ S(12) =[2(100) + (12 – 1)( –4)]
=
3. (a) Let n be the number of terms of the given series.
∵ a = –7, d = –2 – (–7) = 5 and l = T(n) = 103
and T(n) = a + (n – 1)d
∴ 103 = –7 + (n – 1)(5)
n – 1 = 22
n = 23
∴ S(23) =
=
(b) Let N be the number of terms that must be taken.
∵ a = –9, d = –2 – (–9) = 7 and S(n) = 1564
and S(n) =[2a + (N – 1)d]
∴ 1564 =[2(–9) + (N – 1)(7)]
3128 = 7N2 – 25N
7N2 – 25N – 3128 = 0
(N – 23)(7N + 136) = 0
N = 23 or (rejected)
∴ 23 terms of the arithmetic series must be taken.
4. (a) ∵ a = 1, l = 100 and n = 100
∴ 1 + 2 + … + 100 =
= 5050
(b) Let n be the number of terms of the given series.
∵ a = 3, d = 6 – 3 = 3 and l = T(n) = 99
and T(n) = a + (n – 1)d
∴ 99 = 3 + (n – 1)(3)
n = 33
∴ The required sum =
= 1683
(c) The required sum
= sum of integers between 1 and 100 inclusive – sum of integers between 1 and 100 that are multiples of 3
= 5050 – 1683 (from (a) and (b))
=
p. 195
1. (a) ∵ a = 3, R = 2 and n = 6
∴ S(6) =
= 189
(b) ∵ a = 39, R =and n = 8
∴ S(8) =
= 29 520
2. (a) Let n be the number of terms of the given series.
∵ a = 1024, R =and T(n) = aRn – 1 = 8
∴ 8 = 1024
n = 8
∴ S(8) =
= 2040
(b) Let N be the number of terms that must be taken.
∵ a = 1024, R =and S(N) =
∴
N = 14
∴ 14 terms of arithmetic series must be taken.
3. Let the least number of terms of the geometric series be N.
∵ a = 38, and
∴
∵ N is the least number and it is an integer.
∴ N = 10
∴ There should be at least 10 terms.
p. 202
1. (a) ∵ a = 1 and R =
∴
(b) ∵ a = 1, R =
∴
2. (a) = 2222…
= 0.2 + 0.02 + 0.002 + 0.0002 + …
(b) = 0.242 424…
= 0.24 + 0.0024 + 0.000 024 + …
3. (a) C2B2 =C1B1 (mid-pt. theorem)
B2A2 =B1A1 (mid-pt. theorem)
A2C2 =A1C1 (mid-pt. theorem)
∴ Perimeter of △A2B2C2
=´ perimeter of △A1B1C1
=(16 cm)
=
Similarly, perimeter of △A3B3C3
=´ perimeter of △A2B2C2
=(8 cm)
=
(b) Perimeter of △A2B2C2
=´ perimeter of △A1B1C1
=´ 16 cm
Perimeter of △A3B3C3
=´ perimeter of △A2B2C2
=´´ perimeter of △A1B1C1
=´´ 16 cm
=´ 16 cm
∴ Perimeter of △AkBkCk
=´ 16 cm
=
(c) From (b), the perimeters of the triangles formed are in geometric sequence with common ratio.
∴ Sum of the perimeters
= 32 cm
p. 207
1. X1 = 5, X2 = 7, X3 == 6, X4 == 6.5,
X5 == 6.25
2. Y1 = 3, Y2 = 3(2 – 3) = –3, Y 3 = –3[2 – (–3)] = –15,
Y 4 = –15[2 – (–15)] = –255,
Y 5 = –255[2 – (–255)] = –65 535
Exercise
Exercise 14A (p. 159)
Level 1
1. T(1) = 2(1) + 3 =
T(2) = 2(2) + 3 =
T(3) = 2(3) + 3 =
T(4) = 2(4) + 3 =
2. T(1) ==
T(2) =
T(3) =
T(4) =
3. T(1) = 12 – 3 =
T(2) = 22 – 3 =
T(3) = 32 – 3 =
T(4) = 42 – 3 =
4. T(1) =
T(2) =
T(3) =
T(4) =
5. T(1) = (–2)1 – 1 + 3 =
T(2) = (–2)2 – 1 + 3 =
T(3) = (–2)3 – 1 + 3 =
T(4) = (–2)4 – 1 + 3 =
6. T(1) = 32(1) – 1 =
T(2) = 32(2) – 1 =
T(3) = 32(3) – 1 =
T(4) = 32(4) – 1 =
7. (a) 64, 128
(b) ∵ T(1) = 4 = 21 + 1
T(2) = 8 = 22 + 1
T(3) = 16 = 23 + 1
T(4) = 32 = 24 + 1
∴ T(n) =
(c) ∵ T(n) = 2n + 1
∴ T(8) = 28 + 1 = 512
and T(10) = 210 + 1 = 2048
∴ The 8th term and the 10th term of the sequence are 512 and 2048 respectively.
8. (a) 27, 33
(b) ∵ T(1) = 3 = 6(1) – 3
T(2) = 9 = 6(2) – 3
T(3) = 15 = 6(3) – 3
T(4) = 21 = 6(4) – 3
∴ T(n) =
(c) ∵ T(n) = 6n – 3
∴ T(8) = 6(8) – 3 = 45
and T(10) = 6(10) – 3 = 57
∴ The 8th term and the 10th term of the sequence are 45 and 57 respectively.
9. (a) –1, 1
(b) ∵ T(1) = –1 = (–1)1
T(2) = 1 = (–1)2
T(3) = –1 = (–1)3
T(4) = 1 = (–1)4
∴ T(n) =
(c) ∵ T(n) = (–1)n
∴ T(8) = (–1)8 = 1
and T(10) = (–1)10 = 1
∴ The 8th term and the 10th term of the sequence are 1 and 1 respectively.
10. (a)
(b) ∵ T(1) =
T(2) =
T(3) =
T(4) =
∴ T(n) =
(c) ∵ T(n) =
∴ T(8) =
and T(10) =
∴ The 8th term and the 10th term of the sequence are and respectively.
Level 2
11. T(1) =, T(7) == , T(11) ==
12. T(1) =
T(7) =
T(11) =
13. T(1) ==
T(7) =
T(11) =
14. T(1) ==
T(7) =
T(11) =
15. (a) –5, 6
(b) ∵ T(1) = –1 = (–1)1 ´ 1
T(2) = 2 = (–1)2 ´ 2