5ach14(arithmetic And Geometric Sequences And Their Summation)

Certificate Mathematics in Action Full Solutions 5A

14 Arithmetic and Geometric Sequences and their Summation • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Activity

(b) very close to zero

Activity 14.1 (p. 161)

(c) very large

1.

Follow-up Exercise

(a) 2, 2, 2, 2 (b) 3, 3, 3, 3

p. 158

(c) 5, 5, 5, 5

2.

1.

(a) T(1) = 5(1) = 5

(d) –5, –5, –5, –5

T(2) = 5(2) = 10

They are equal.

T(3) = 5(3) = 15 T(4) = 5(4) = 20

Activity 14.2 (p. 184) 1.

(b) T(1) = 12 – 1 = 0

(a) S(10) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

T(2) = 22 – 1 = 3

(b) S(10) = 55 2.

T(3) = 32 – 1 = 8 T(4) = 42 – 1= 15

No

Activity 14.3 (p. 198) 1 .

Range of R

1

R

very large

Rn 256 (or any other reasonable answers) –32 768 (or any other reasonable answers) very large or very small

8

1 (or any 256

n 8

R < –1

–2 (or any other reasonable answers)

15

1 2 –1 < R < 1

(or any other reasonable answers)

15

very large

R>1

2.

2 (or any other reasonable answers)

(a) very large or very small

8 15 very large

other reasonable answers) 1 (or 32 768 any other reasonable answers) very close to zero 256 (or any other reasonable answers) 32 768 (or any other reasonable answers) very large

1 1  (c) T(1) =   = 3 3   1  3

2

T(2) =   = 3

1  3

T(3) =   = 4

1  3 

T(4) =   = 2.

1 9 1 27 1 81

(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) = 5n −4 (iii) ∵ ∴ and ∴

T(n) = 5n – 4 T(10) = 5(10) – 4 = 46 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 n− 1 ∴ T(n) = 3

Certificate Mathematics in Action Full Solutions 5A

(iii) ∵ ∴ and ∴

(c) (i)

T(n) = 3n – 1 T(10) = 310 – 1 = 39 T(15) = 315 – 1 = 314 The 10th term and the 15th term of the sequence are 39 and 314 respectively.

= 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.

1 1 , 32 64

1 1 = 1 2 2 1 1 T(2) = = 2 4 2 1 1 T(3) = = 3 8 2 1 1 = 4 T(4) = 16 2

2.

(ii) ∵ T(1) =

∴ T(n) =

(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) = 54

1 2n

1 2n 1 1 ∴ T(10) = 10 = 2 1024 1 1 and T(15) = 15 = 2 32 768

(iii) ∵ T(n) =

(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) 16 =−

∴ The 10th term and the 15th term of the sequence are

1 1024

and

1 32 768

respectively.

(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 =8

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 ≠ 2 ∴ It is not an arithmetic sequence. (d) T(2) – T(1) = log 4 – log 2 = 2 log 2 – log 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 = 35

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.

14

Arithmetic and Geometric Sequences and their Summation

(b) T(n) = a + (n – 1)d = –2 + (n – 1)(3) 5 +3n =− (c) ∵ T(k) = 73 ∴ –5 + 3k = 73 k = 26 4.

(a) T(n) = a + (n – 1)d = –24 + (n – 1)(5) 29 +5n =−

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.

6+ y   x= 2 ∴ y + 22 18 = 2 

(b) T(7) = – 29 + 5(7) =6 T(12) = – 29 + 5(12) = 31

x=

29 m> 5

p. 170 1.

(a)

−10 + ( −2) 2 137 + 27 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.

80

6 +14 2

T ( 2) 4 = =2 T (1) 2 T (3) 6 3 = = ≠2 T ( 2) 4 2 ∴ It is not a geometric sequence.

(b)

= 82 2.

2)

p. 176

= –6 (b) Arithmetic mean =

.........(

= 10

1.

(a) Arithmetic mean =

1)

From (2), we have 36 = y + 22 y = 14 By substituting y = 14 into (1), we have

(c) ∵ The mth term is the first positive term of the sequence. ∴ T(m) > 0 i.e. – 29 + 5m > 0

∵ m is the number of terms, it must be an integer. ∴ m=6

.........(

∴

(c)

T ( 2) −4 1 = = T (1) −8 2 T (3) −2 1 = = T (2) −4 2 T (4) −1 1 = = T (3) −2 2 It is a geometric sequence with common ratio

T ( 2) 0.22 11 = = T (1) 0.2 10 T (3) 0.222 111 11 = = ≠ T ( 2) 0.22 110 10 ∴ It is not a geometric sequence.

(d)

T ( 2) log 9 = =2 T (1) log 3 T (3) log 27 3 = = ≠2 T (2) log 9 2

1 . 2

Certificate Mathematics in Action Full Solutions 5A

∴ It is not a geometric sequence. 2.

∵ a = –8, R =

(a) Let a and R be the first term and the common ratio respectively. ∵ a = 1 and R =

n −1

1  ∴ T(n) = −8  2 

2 =2 1

= −2 4−n

∴ T(n) = aRn – 1 = 1(2)n – 1 = 2n – 1 ∴ T(6) = 26 – 1 = 32

(b) T(4) = –24 – 4 1 =− T(6) = –24 – 6 1 =− 4

(b) Let a and R be the first term and the common ratio respectively. ∵ a = 3 and R =

9 =3 3

4.

∴ T(n) = aRn – 1 = 3(3)n – 1 = 3n ∴ T(6) = 36 = 729

(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

(c) Let a and R be the first term and the common ratio respectively. ∵ a = –3 and R =

a=

6 = –2 −3

3 respectively.

3 (–2)n 2 3 ∴ T(6) = (–2)6 2

(b) T(n) = aRn – 1

=

=

−64 1 =− 128 2

∴ T(n) = aRn – 1



 



∴ T(6) = –256  −



1.

n −1

1  2

6

1  2

(a) Let a and R be the first term and the common ratio respectively.

(a) Geometric mean = 5 ×45 = 15 (b) Geometric mean = −12 ×( −147 ) = 42

n

4 =−

3.

p. 181

1  2

= –256  −

n− 3

(c) T(7) + T(9) = 37 – 3 + 39 – 3 = 810

(d) Let a and R be the first term and the common ratio respectively.



1 ⋅ 3n – 1 9

=3

= 96

= 128  −

1 9

∴ The first term and the common ratio are

∴ T(n) = aRn – 1 = –3(–2)n – 1

∵ a = 128 and R =

1 and T(n) = aRn – 1 2

2.

(a) Let R be the common ratio of the geometric sequence to be formed. The geometric sequence formed is:

1 1 1 27 , R, R 2 , 2 2 2 16 ∵ The 4th term is also given by

1 3 R. 2

1 and 9

14

∴

Arithmetic and Geometric Sequences and their Summation

(b) ∵ a = 3, d = 8 – 3 = 5 and n = 12

1 3 27 R = 2 16 3 R= 2

∴ S(12) =

= 366

∴ The two required geometric means are

3 and 4

(c) ∵ a = 28, d = 26 – 28 = –2 and n = 15 ∴ S(15) =

9 . 8 2.

(a) ∵ a = –5, l = 9 and n = 8 ∴ S(8) =

3 3 3 2 3 3 2 , r, r , r , 8 8 8 8 27 ∵ The 5th term is also given by

3 4 r. 8

3 4 2 r = 8 27 2 r=± 3

(b) ∵ a = 100, d = –4 and n = 12 ∴ S(12) =

3.

(b) Let N be the number of terms that must be taken. ∵ a = –9, d = –2 – (–9) = 7 and S(n) = 1564

N [2a + (N – 1)d] 2 N 1564 = [2(–9) + (N – 1)(7)] 2

and

(a) 1 + 3 + 5 + 7 + 9

3128 = 7N2 – 25N 7N – 25N – 3128 = 0 (N – 23)(7N + 136) = 0 2

(a) 2 + 4 + 8 + 16 + 32 + 64

N = 23 or −

(a) 1 + 4 + 9 + 16 + 25 + 36 + 49

(a) ∵ a = 1, d = 5 – 1 = 4 and n = 10 ∴ S(10) =

10 [2(1) + (10 – 1)(4)] 2

= 190

136 (rejected) 7

∴ 23 terms of the arithmetic series must be taken. 4.

p. 188

80

S(n) =

∴

(b) 140

1.

23 ( −7 +103 ) 2

= 1104

(b) 126 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) =

∵ 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) 2 x + 6x + 9 = x2 + 9x 3x = 9 x=3

(b) 25 2.

12 [2(100) + (12 – 1)( –4)] 2

= 936

p. 183 1.

8( −5 +9) 2

= 16

1 ∴ The three required geometric means are , 4 1 1 1 1 1 and or − , and − . 6 9 9 4 6 3.

15 [2(28) + (15 – 1)( –2)] 2

= 210

(b) Let r be the common ratio of the geometric sequence to be formed. The geometric sequence formed is:

∴

12 [2(3) + (12 – 1)(5)] 2

(a) ∵ a = 1, l = 100 and n = 100 ∴ 1 + 2 + … + 100 =

100 (1 +100 ) 2

= 5050 (b) Let n be the number of terms of the given series. ∵ a = 3, d = 6 – 3 = 3 and l = T(n) = 99

Certificate Mathematics in Action Full Solutions 5A

and T(n) = a + (n – 1)d ∴ 99 = 3 + (n – 1)(3) n = 33 ∴ The required sum =

33 (3 + 99 ) 2

∴

= 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)) = 3367

N

1 1 2048  = 8  2 N = 14 ∴ 14 terms of arithmetic series must be taken. 3.

Let the least number of terms of the geometric series be N.

37 1 = and 38 3 a (1 − R N ) S(N ) = > 9841 1−R   1 N  8 3 1−     3    > 9841 1 1−

p. 195 1.

∵ a = 38,

(a) ∵ a = 3, R = 2 and n = 6 ∴ S(6) =

3( 2 6 −1) 2 −1

= 189

(b) ∵ a = 39, R =

∴

1 and n = 8 3

  1 8  39 1 −      3   S(8) = 1 1− 3

∴

3 2

N

N

1 1 −  > − 19 683  3

512 1 = and T(n) = aRn – 1 = 8 1024 2

N

1    3  1 3N 3N N

n −1

∴

1  8 = 1024   2

n −1

1 1  =  128  2  n=8

∴

  1 8  1024 1 −      2   S(8) = 1 1− 2

  1 N  1 −    > 9841   3    19 682 1 1−   > 19 683  3

(a) Let n be the number of terms of the given series. ∵ a = 1024, R =

R=

3

9

= 29 520 2.

  1 N  10241 −      2   7 2047 = 8 1 − 12

∵ N is the least number and it is an integer. ∴ N = 10 ∴ There should be at least 10 terms.

p. 202

= 2040 (b) Let N be the number of terms that must be taken.

1 and S(N) = 2 a (1 − R N ) 7 = 2047 1−R 8

1 19 683 1 < 9 3 > 39 >9 <

1.

1 

 (a) ∵ a = 1 and R =  2  = 1 1

∵ a = 1024, R =

∴

S (∞) =

1 1−

1 2

2

=2

14

Arithmetic and Geometric Sequences and their Summation

1 × perimeter of △A2B2C2 2 1 1 = × × perimeter of △A1B1C1 2 2 1 1 = × × 16 cm 2 2

− 1   3  = −1 1 3

=

(b) ∵ a = 1, R = 

∴

2.

(a)

S (∞) =

1 3 =  1 4 1 − −   3

2

1  × 16 cm 2

 = 2222… 0.2 = 0.2 + 0.02 + 0.002 + 0.0002 + …

=

0.2 = 1 −0.1 2 = 9

∴ Perimeter of △AkBkCk k −1

1   2 

=

5−k cm =2

4  = 0.242 424… (b) 0.2 = 0.24 + 0.0024 + 0.000 024 + …

(c) From (b), the perimeters of the triangles formed are in geometric sequence with common ratio

0.24 = 1 −0.01 8 = 33

3.

1 C 1B 1 2 1 B 2A 2 = B 1A 1 2 1 A 2C 2 = A 1C 1 2

(a) C2B2 =

∴ Sum of the perimeters

=

16 1−

1

1 . 2

cm

2

= 32 cm (mid-pt. theorem) (mid-pt. theorem)

p. 207 1.

(mid-pt. theorem)

X1 = 5, X2 = 7, X3 = X5 =

∴ Perimeter of △A2B2C2

1 × perimeter of △A1B1C1 2 1 = (16 cm) 2 =

= 8 cm Similarly, perimeter of △A3B3C3

1 × perimeter of △A2B2C2 2 1 = (8 cm) 2 =

2.

6. 5 + 6 = 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 14A (p. 159) Level 1 1.

T(1) = 2(1) + 3 = 5 T(2) = 2(2) + 3 = 7 T(3) = 2(3) + 3 = 9 T(4) = 2(4) + 3 = 11

(b) Perimeter of △A2B2C2

1 × perimeter of △A1B1C1 2 1 = × 16 cm 2 =

Perimeter of △A3B3C3

7 +5 6 +7 = 6, X4 = = 6.5, 2 2

Exercise

= 4 cm

80

× 16 cm

2.

T(1) =

1 = 1

1 T(2) = 2 1 T(3) = 3

1

Certificate Mathematics in Action Full Solutions 5A

∴ 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.

1 T(4) = 4

3.

2 T(1) = 12 – 3 = −

T(2) = 22 – 3 = 1 T(3) = 32 – 3 = 6 T(4) = 42 – 3 = 13 4.

5.

T(1) =

3(1) −2 1 = 9 9

T(2) =

3( 2) −2 4 = 9 9

T(3) =

3(3) −2 7 = 9 9

T(4) =

3( 4) −2 10 = 9 9

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 1) n ∴ 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.

6 7 , 7 8

10. (a)

T(1) = (–2)1 – 1 + 3 = 4 T(2) = (–2)2 – 1 + 3 = 1

2 3 3 T(2) = 4 4 T(3) = 5 5 T(4) = 6

(b) ∵ T(1) =

T(3) = (–2)3 – 1 + 3 = 7 5 T(4) = (–2)4 – 1 + 3 = −

6.

T(1) = 32(1) – 1 = 3 T(2) = 32(2) – 1 = 27 T(3) = 32(3) – 1 = 243 T(4) = 32(4) – 1 = 2187

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 n+ 1 ∴ T(n) = 2 (c) ∵ ∴ and ∴

8.

∴ T(n) =

(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) = 6n −3 (c) ∵ T(n) = 6n – 3

n +1 n +2

n +1 n +2 8 +1 9 = ∴ T(8) = 8 + 2 10 10 +1 11 = and T(10) = 10 + 2 12

(c) ∵ T(n) =

T(n) = 2n + 1 T(8) = 28 + 1 = 512 T(10) = 210 + 1 = 2048 The 8th term and the 10th term of the sequence are 512 and 2048 respectively.

(a) 27, 33

1 +1 1+2 2 +1 = 2 +2 3 +1 = 3 +2 4 +1 = 4 +2 =

∴ The 8th term and the 10th term of the sequence are

9 11 and respectively. 10 12

Level 2 11. T(1) = 243

31 1 37 311 = , T(7) = = 3 , T(11) = = 729 243 729 729

14

12. T(1) = T(7) =

2(1) 2 −1 1 = 1 +1 2

Arithmetic and Geometric Sequences and their Summation

(b) ∵ T(1) =

1 12 = 4 (1 +1) 2

T(2) =

4 22 = 9 ( 2 +1) 2

T(3) =

9 32 = 16 (3 +1) 2

T(4) =

16 42 = 25 ( 4 +1) 2

2(7) 2 −1 97 = 7 +1 8

T(11) =

2(11 ) 2 −1 241 = 11 +1 12

(1 −1)(1 −2) 0 13. T(1) = = 1 T(7) =

(7 −1)( 7 −2) 30 = 7 7

T(11) =

(11 −1)(11 −2) 90 = 11 11

∴ T(n) =

n2 ( n +1) 2

Exercise 14B (p. 166) Level 1

1

14. T(1) = T(7) =

3 = 13

3

1.

T(2) – T(1) = 12 – 16 = –4 T(3) – T(2) = 8 – 12 = –4 T(4) – T(3) = 4 – 8 = –4 ∴ It is an arithmetic sequence with common difference – 4.

2.

T(2) – T(1) =

37 2187 = 3 7 343

T(11) =

311 177 147 = 3 11 1331

15. (a) –5, 6 (b) ∵ T(1) = –1 = (–1)1 × 1 T(2) = 2 = (–1)2 × 2 T(3) = –3 = (–1)3 × 3 T(4) = 4 = (–1)4 × 4 1) n n ∴ T(n) = (−

3.

16. (a) log 80, log 160 (b) ∵ T(1) = log 5 = log (5 ⋅ 21 – 1) T(2) = log 10 = log (5 ⋅ 22 – 1) T(3) = log 20 = log (5 ⋅ 23 – 1) T(4) = log 40 = log (5 ⋅ 24 – 1) n− 1 ∴ T(n) = log (5 ⋅2 )

18. (a)

25 36 , 36 49

80

5 1 1 − = 6 3 2 4 5 1 T(3) – T(2) = − = 3 6 2 11 4 1 − = T(4) – T(3) = 6 3 2 T(2) – T(1) =

∴ It is an arithmetic sequence with common difference

1 . 2 4.

17. (a) 30, 42 (b) ∵ T(1) = 2 = 1 × (1 + 1) T(2) = 6 = 2 × (2 + 1) T(3) = 12 = 3 × (3 + 1) T(4) = 20 = 4 × (4 + 1) 1) ∴ T(n) = n( n +

2 − 1 = 2 −1

T(3) – T(2) = 3 − 2 ≠ 2 −1 ∴ It is not an arithmetic sequence.

2 1 1 − = 3 2 6 3 2 1 1 ≠ T(3) – T(2) = − = 4 3 12 6 T(2) – T(1) =

∴ It is not an arithmetic sequence. 5.

(a) ∵ a = 5 and d = 2 and T(n) = a + (n – 1)d ∴ T(n) = 5 + (n – 1)(2) = 3 +2 n (b) T(10) = 3 + 2(10) = 23

6.

(a) ∵ a = –3 and d = 9 and T(n) = a + (n – 1)d 12 +9n ∴ T(n) = –3 + (n – 1)(9) = −

Certificate Mathematics in Action Full Solutions 5A

∴

–100 + 7k = 5 k = 15 ∴ There are 15 terms in the sequence.

(b) T(10) = –12 + 9(10) = 78 7.

3 and d = –2 2

(a) ∵ a =

and T(n) = a + (n – 1)d ∴ T(n) =

(b) T(10) =

8.

7 3 −2n + (n – 1)(–2) = 2 2

33 7 – 2(10) = − 2 2

(a) ∵ a = –2 and d =

3 2

and T(n) = a + (n – 1)d ∴ T(n) = –2+ (n – 1)(

(b) T(10) =

9.

1 3 ) = (3n −7) 2 2

23 1 [3(10) – 7] = 2 2

d=6–1=5 T(n) = a + (n – 1)d = 1 + (n – 1)(5) = 5n −4

2 10. d = 17 – 19 = −

T(n) = a + (n – 1)d = 19 + (n – 1)(–2) = 21 −2n 11. d = log 9 – log 3 = 2 log 3 – log 3 = log 3 T(n) = a + (n – 1)d = log 3 + (n – 1)(log 3) =n log

3

12. d = (2a – 3c) – (a – 2c) = a −c T(n) = a + (n – 1)d = (a – 2c) + (n – 1)(a – c) = − c +n ( a − c)

13. Let a and d be the first term and the common difference respectively. ∵ a = 7 and d = 11 – 7 = 4 ∴ T(n) = a + (n – 1)d = 7 + (n – 1)(4) = 3 + 4n Let 83 be the kth term. i.e. T(k) = 83 ∴ 3 + 4k = 83 k = 20 ∴ There are 20 terms in the sequence. 14. Let a and d be the first term and the common difference respectively. ∵ a = –93 and d = –86 – (–93) = 7 ∴ T(n) = a + (n – 1)d = –93 + (n – 1)(7) = –100 + 7n Let 5 be the kth term. i.e. T(k) = 5

15. Let a and d be the first term and the common difference respectively. T(2) = a + d = 18 ……(1) T(6) = a + 5d = 30 ……(2) (2) – (1), 4d = 12 d=3 By substituting d = 3 into (1), we have a + 3 = 18 a = 15 ∴ T(n) = a + (n – 1)d = 15 + (n – 1)(3) = 12 +3n 16. Let a and d be the first term and the common difference respectively. T(4) = a + 3d = –2 ……(1) T(9) = a + 8d = –32 ……(2) (2) – (1), 5d = –30 d = –6 By substituting d = –6 into (1), we have a + 3(–6) = –2 a = 16 ∴ T(n) = a + (n – 1)d = 16 + (n – 1)(–6) = 22 −6n 17. Let a and d be the first term and the common difference respectively. T(3) = a + 2d = 60 ……(1) T(7) = a + 6d = 40 ……(2) (2) – (1), 4d = –20 d = –5 By substituting d = –5 into (1), we have a + 2(–5) = 60 a = 70 ∴ T(n) = a + (n – 1)d = 70 + (n – 1)(–5) = 75 −5n 18. Let a and d be the first term and the common difference respectively. T(3) = a + 2d = 82 ……(1) T(10) = a + 9d = 250 ……(2) (2) – (1), 7d = 168 d = 24 By substituting d = 24 into (1), we have a + 2(24) = 82 a = 34 ∴ T(n) = a + (n – 1)d = 34 + (n – 1)(24) = 10 +24 n 19. (a) 1, 2; 2, 4; 3, 6 (or any other reasonable answers) (b) –1, 1; –2, 2; –3, 3 (or any other reasonable answers)

Level 2 20. Let a and d be the first term and the common difference respectively. a = –101,d = –98 – (–101) = 3 ∴ T(n) = a + (n – 1)d = –101 + (n – 1)(3)

14

= –104 + 3n ∵ The kth term is the first positive term of the sequence. ∴ T(k) > 0 i.e. –104 + 3k > 0

104 k> 3 ∵ k is the number of terms, it must be an integer. ∴ k = 35 21. Let a and d be the first term and the common difference respectively. a = 999, d = 992 – 999 = –7 ∴ T(n) = a + (n – 1)d = 999 + (n – 1)(–7) = 1006 – 7n Let the kth term be the last positive term. ∴ T(k) > 0 i.e. 1006 – 7k > 0 k<

1006 7

∵ k is the number of terms, it must be an integer. ∴ k = 143 ∴ There are 143 positive terms. 22. Let a, (a + d) and (a + 2d) be the interior angles of the triangle respectively. ∵ a = 15° ∴ The other two angles are 15° + d and 15° + 2d. 15° + (15° + d) + (15° + 2d) = 180° (∠ s sum of triangle) d = 45° ∴ The largest angle is 105°. 23. The multiples of 13 form an arithmetic sequence with a = 13 and d = 13. ∴ T(n) = a + (n – 1)d = 13 + (n – 1)(13) = 13n Consider the term that is less than 1000. i.e. T(n) < 1000 13n < 1000 n<

1000 13

∵ Among the terms that is less than 1000, the 76th term is the greatest. ∴ The greatest integer which is a multiple of 13 and less than 1000 is 13(76) = 988. 24. (a) Let a and d be the first term and the common difference respectively. T(5) = 4T(1) a + 4d = 4a 3a – 4d = 0 ……(1) T(6) = 2T(3) – 1 a + 5d = 2(a + 2d) – 1 a–d=1 ……(2) (2) × 4 – (1), a = 4 By substituting a = 4 into (2), we have 4–d=1

80

Arithmetic and Geometric Sequences and their Summation

d=3 ∴ T(n) = a + (n – 1)d = 4 + (n – 1)(3) = 1 +3n (b)

T(m) – T(25) = 81 [1 + 3m] – [1 + 3(25)] = 81 m = 52

25. Let (a – d) cm, a cm and (a + d) cm be the lengths of the sides of the right-angled triangle respectively. ∵ The perimeter is 27 cm. ∴ (a – d) + a + (a + d) = 27 a=9 (a – d)2 + a2 = (a + d)2 (Pyth. theorem) a2 – 2ad + d2 + a2 = a2 + 2ad + d2 a2 = 4ad ( a ≠ 0) a = 4d d=

1 a………(1) 4

By substituting a = 9 into (1), we have d = 2.25. ∴ The lengths of the three sides of the triangle are 6.75 cm, 9 cm and 11.25 cm. 26. (a) ∵ T(9) = 3T(4) ∴ a + 8d = 3(a + 3d) a + 8d = 3a + 9d d a=− 2 ∴ T(n) = a + (n – 1)d = −

d + (n – 1)d 2

 3  = − 2 +nd   (b) ∵

T(k) = 5T(5)

 3   3   − + k d = 5 − + 5 d  2   2  k =19 27. Let the three terms be a – d, a and a + d respectively. ∵ The sum is 15. ∴ (a – d) + a + (a + d) = 15 a=5 ∵ The product is 80. ∴ (a – d)(a)(a + d) = 80 ………(1) By substituting a = 5 into (1), we have (5 – d)(5)(5 + d) = 80 d=±3 When d = –3, a – d = 5 – (–3) = 8 and a + d = 5 + (–3) = 2 When d = 3, a – d = 5 – 3 = 2 and a + d = 5 + 3 = 8 ∴ The terms are 2, 5, 8. 28. Let a and d be the first term and the common difference

Certificate Mathematics in Action Full Solutions 5A

respectively. aT(6) = –38 a(a + 5d) = –38 T(3) + T(4) = 17 (a + 2d) + (a + 3d) = 17

………(1)

17 − 2a d= 5

………(2)

By substituting (2) into (1), we have

 17 − 2a   ] = –38 5  

a[a + 5 

a2 – 17a – 38 = 0 (a – 19)(a + 2) = 0 a = 19 or a = –2(rejected) By substituting a = 19 into (2), we have

17 − 2(19 ) 21 =− d= 5 5 ∴ The first term and the common difference are 19 and

−

21 respectively. 5

29. (a) T(2) – T(1) = log 10k2 – log 10k = log

10 k 2 10 k

= log k T(3) – T(2) = log 10k3 – log 10k2

10 k 3 = log 10 k 2 = log k T(4) – T(3) = log 10k4 – log 10k3 = log ∴ ∵ ∴

∴

10 k 4 10 k 3

= log k It is an arithmetic sequence. a = log 10k and d = log k T(n) = a + (n – 1)d = log 10k + (n – 1)(log k) = 1 + n log k The general term is 1 + n log k.

(b) T(2) – T(1) = log (10k2)2 – log (10k)2 = 2(log 10k2 – log 10k) = 2 log k T(3) – T(2) = log (10k3)2 – log (10k2)2 = 2(log 10k3 – log 10k2) = 2 log k T(4) – T(3) = log (10k4)2 – log (10k3)2 = 2(log 10k4 – log 10k3) = 2 log k ∴ It is an arithmetic sequence. ∵ a = log (10k)2 and d = 2 log k ∴ T(n) = a + (n – 1)d

= log (10k)2 + (n – 1)(2 log k) = 2 + 2n log k ∴ The general term is 2 + 2n log k. 30. (a) Let the three terms be a , a + d and a + 2d respectively. a(a + d)(a + 2d) = a + (a + d) + (a + 2d) a(a + d)(a + 2d) = 3(a + d) [a(a + 2d) – 3] (a + d) = 0 a(a + 2d) – 3 = 0

or

d a=−

2

a + 2ad – 3 = 0 a=

−2d ± ( 2d ) 2 −4( −3) 2(1)

a = −d ± d 2 +3 (b) If a = −d + d 2 +3 , then a = −1 + 12 +3 = 1. ∴ The terms are 1, 2, 3. If a = −d − d 2 +3 , then a = −1 − 12 +3 = –3. ∴ The terms are –3, –2, –1. If a = –d, then a = –1. ∴ The terms are –1, 0, 1. 31. Let the three integers be a – d, a and a + d respectively, and their sum be k. (a – d) + a + (a + d) = k 3a = k………(1) (a – d)(a)(a + d) = 11k………(2) By substituting (1) into (2), we have (a – d)(a)(a + d) = 11(3a) (a – d)(a + d) = 33

( a ≠ 0)

∵ a and d are integers. ∴ Possible solutions are:

 a+ d = 3 3  a+ d = 1 1  a+ d = 3 or or or     a− d = 1  a− d = 3  a− d = 1 1  a+ d = 1   a− d = 3 3

14

i.e.

a= 1 7 a= 7 a= 7 a= 1 7 or or   or   d = 1 6  d = 4  d = − 4  d = − 1 6

Arithmetic and Geometric Sequences and their Summation

1, 0.5, 2, 3.5, 5 and 6.5. 4.

Let d be the common difference of the arithmetic sequence. ∵ a = 18 and T(5) = 36 ∴ 36 = 18 + 4d d=

If a = 17 and d = 16, the three integers are 1, 17, 33. If a = 7 and d = 4, the three integers are 3, 7, 11. If a = 7 and d = –4, the three integers are 11, 7, 3. If a = 17 and d = –16, the three integers are 33, 17, 1. ∴ The three integers are 1, 17, 33 or 3, 7, 11.

9  = 22 .5 2

∴ a = 18 + 

9  = 27 2

b = 18 + 2 

Exercise 14C (p. 170)

9  = 31 .5 2

c = 18 + 3 

Level 1 1.

(a) Arithmetic mean =

−8 + 4 = −2 2

5.

(b) Arithmetic mean =

7 + 25 = 16 2

Level 2

42 +104 = 73 (c) Arithmetic mean = 2

6.

2.

∵

3.

(a) Let d1 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: –4, –4 + d1, –4 + 2d1, 8 ∵ The 4th term is also given by –4 + 3d1. ∴ –4 + 3d1 = 8 d1 = 4 ∴ The two required arithmetic means are 0 and 4. (b) Let d2 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: –4, –4 + d2, –4 + 2d2, –4 + 3d2, 8 ∵ The 5th term is also given by –4 + 4d2. ∴ –4 + 4d2 = 8 d2 = 3 ∴ The three required arithmetic means are –1, 2 and 5. (c) Let d3 be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: –4, –4 + d3, –4 + 2d3, –4 + 3d3, –4 + 4d3, –4 + 5d3, –4 + 6d3, –4 + 7d3, 8 ∵ The 9th term is also given by –4 + 8d3. ∴ –4 + 8d3 = 8 d3 = 1.5 ∴ The seven required arithmetic means are –2.5, –

1, 2, 4, 5; –1, 1, 5, 7; –3, 0, 6, 9 (or any other reasonable answers)

∵ 2x + 1 is the arithmetic mean between x and 14. ∴ 2x + 1 =

x +14 2

4x + 2 = 14 + x x= 4 ∴ a = 4 and d = (2x + 1) – x = x + 1 = 5 T(n) = a + (n – 1)d = 4 + (n – 1)(5) = 5n −1

8 +x = 25 2 x = 42

80

9 2

7.

∵ x is the arithmetic mean between 8 and y, and 21 is the arithmetic mean between y and 26.

8+ y   x= 2 ∴ y + 26 21 = 2 

.........(

1)

.........(

2)

From (2), we have 42 = y + 26 y = 16 By substituting y = 16 into (1), we have x= 8.

8 +16 =12 2

(a) Let d be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: 1, 1 + d, 1 + 2d, 1 + 3d, p ∵ The 5th term is also given by 1 + 4d. ∴ 1 + 4d = p d=

p −1 4

∴ The three required arithmetic means are

Certificate Mathematics in Action Full Solutions 5A

p + 3 p +1 3 p +1 , and . 4 2 4 (b) common difference =

∴ It is not a geometric sequence.

p −1 (from (a)) 4

2.

 p −1    4 

T(n) = a + (n – 1)d = 1 + (n – 1) 

( n −1)( p −1) 4 ( n −1)( p −1) ∴ The general term is 1 + . 4 =1+

9.

(a) Arithmetic mean =

∴ It is a geometric sequence with common ratio –4. 3.

( a − d ) + ( a + d ) 2a = =a 2 2 4.

10. (a) (i) The arithmetic sequence formed is: x, x + d1, x + 2d1, x + 3d1, …, x + nd1, y ∴ The arithmetic means are x + d1, x + 2d1, x + 3d1, …, x + nd1

(b) (i) The arithmetic sequence formed is: x, x + d2, x + 2d2, x + 3d2, …, x + md2, y ∴ The arithmetic means are x + d2, x + 2d2, x + 3d2, …, x + md2

T ( 2) −10 = = –5 T (1) 2 T (3) −50 = = 5 ≠ −5 T ( 2) −10 ∴ It is not a geometric sequence.

(b) By substituting a = 132 and d = 92 into (a), we have 2 arithmetic mean = 13

(ii) ∵ The (n +2)th term is also given by x + (n +1)d1. ∴ x + (n +1)d1 = y y −x d1 = n +1

T ( 2) −8 = = –4 T (1) 2 T (3) 32 = = –4 T ( 2) −8 T ( 4) −128 = = –4 T (3) 32

T ( 2) 0.1 = = 0.5 T (1) 0.2 T (3) 0.05 = = 0.5 T ( 2) 0.1 T ( 4) 0.025 = = 0.5 T (3) 0.5 ∴ It is a geometric sequence with common ratio 0.5.

5.

(a) ∵ a = 2 and R = 3 and T(n) = aRn – 1 n− 1 ∴ T(n) = 2 ⋅3 (b) T(8) = 2 ⋅ 38−1 = 4374

6.

(ii) ∵ The (m +2)th term is also given by x + (m +1)d2. ∴ x + (m +1)d2 = y y −x d2 = m +1

(or 2 ⋅37 )

(a) ∵ a = –3 and R =

1 and T(n) = aRn – 1 3 n −1

1  =( −3)  ∴ T(n) 3  =−32−n

(b) T(8) = –32 – 8 1  1  = − 729 or − 36   

y −x y −x : = ( m +1) : (n +1) (c) d1 : d2 = n +1 m +1 Exercise 14D (p. 176) Level 1 1.

T ( 2) 10 = =2 T (1) 5 T (3) 15 3 = = ≠2 T ( 2) 10 2

7.

(a) ∵ a = –4 and R = −

3 and T(n) = aRn – 1 2

∴ T(n) = n− 1

 3 ( −4)−   2

(or ( −1) n ⋅ 23−n ⋅3n−1 )

14

8−1

 3 (b) T(8) = ( −4)− 2    8.

=

2187 32

 37   or 25    

n–1 2 and R = − 2 and T(n) = aR

(a) ∵ a =

2 ( − 2 ) n−1

∴ T(n) =

n  2 )n  1) n−1 ⋅2 2 or ( − 

=( − 1) n−1 (

   

Arithmetic and Geometric Sequences and their Summation

= 37 – n Let 1 be the kth term. i.e. T(k) = 1 ∴ 37 – k = 1 ∴ 7–k=0 k=7 ∴ There are 7 terms in the sequence. 14. Let a and R be the first term and the common ratio respectively. a = 1.5, R =

1 1)8− ( (b) T(8) = ( −

9.

2 )8 =− 16

Let a be the first term. ∵ a = 2 and R =

6 =3 2

∴ T(n) = aRn – 1 n− 1 = 2 ⋅3 10. Let a be the first term. 2 1 = ∵ a = 6 and R = 6 3 ∴ T(n) = aR

n −1

1  = 6  3  2 = n−2 3

T(2) = aR =

1 4

………(1)

T(7) = aR6 = 8 (2) ÷ (1),

………(2) R5 = 32 R=2 By substituting R = 2 into (1), we have

1 4 1 a= 8

a(2) =

11. Let a be the first term. ∵ a = 2 2 and R =

2 3 2 2

=

3 2

∴ T(n) = aRn – 1 =2

∴ T(n) = aRn – 1 = 1.5(–2)n – 1 Let 96 be the kth term. i.e. T(k) = 96 ∴ 1.5(–2)k – 1 = 96 (–2)k – 1 = 64 (–2)k – 1 = (–2)6 ∴ k–1=6 k=7 ∴ There are 7 terms in the sequence. 15. Let a and R be the first term and the common ratio respectively.

n–1

 2  

−3 = –2 1.5

3 2

n− 1

   

n n− 1  4−  2 or  ⋅3 2  2   

∴ T(n) = aRn – 1 =

1 ⋅ 2n – 1 8

=2

n −4

12. Let k be the first term. ∵ k = a and R =

a 2r = ar ar

∴ T(n) = kRn – 1 n n− 1 = a ⋅r

243 1 = 729 3

∴ T(n) = aRn – 1 n −1

1   3 

= 729 

80

1 27 1 R=− 3 1 By substituting R = − into (1), we have 3 R3 = −

(2) ÷ (1),

13. Let a and R be the first term and the common ratio respectively. a = 729, R =

16. Let a and R be the first term and the common ratio respectively. T(3) = aR2 = –27 ………(1) T(6) = aR5 = 1 ………(2)

2

 1  = –27  3

a −

Certificate Mathematics in Action Full Solutions 5A

a = –243 ∴ T(n) = aRn – 1 n −1

 1 = − 243  −   3 n 6 −n = ( −1) ⋅3

17. Let a and R be the first term and the common ratio respectively. T(2) = aR = 5 ………(1) T(7) = aR6 = 160 ………(2) (2) ÷ (1), R5 = 32 R=2 By substituting R = 2 into (1), we have a(2) = 5 a=

5 2

5 ⋅ 2n – 1 2

= 5 ⋅2

T(7) = aR6 =

n −2

1 3

………(2)

(2) ÷ (1),

R3 = R=

1 729

1 9

1 By substituting R = into (1), we have 9

………(2)

1 16 1 R=± 2 1 By substituting R = into (1), we have 2 R4 =

2

1  =1 2

a

a=4 ∴ T(n) = aRn – 1 n −1

1   2

=4

= 23 – n

1  = 243 9

a

By substituting R = −

a = 2187 ∴ T(n) = aRn – 1 n −1

1   9 

= 2187 ⋅  9 −2 n

=3

19. (a) 2, 4; 3, 9; 4, 16 (or any other reasonable answers) (b)

1 16

(2) ÷ (1),

18. Let a and R be the first term and the common ratio respectively. T(2) = aR = 243 ………(1) T(5) = aR4 =

………(1) ………(2) R4 = 16 R=±2 By substituting R = 2 into (1), we have a(2) = 6 a=3 ∴ T(n) = aRn – 1 = 3 ⋅ 2n–1 By substituting R = –2 into (1), we have a(–2) = 6 a = –3 ∴ T(n) = aRn – 1 = –3(–2)n – 1 ∴ The general term is 3 ⋅ 2n – 1 or –3(–2)n – 1. 21. Let a and R be the first term and the common ratio respectively. T(3) = aR2 = 1 ………(1)

∴ T(n) = aRn – 1 =

T(2) = aR = 6 T(6) = aR5 = 96 (2) ÷ (1),

1 1 1 , 3; , 2; − , –3 3 2 3 (or any other reasonable answers)

Level 2 20. Let a and R be the first term and the common ratio respectively.

1 into (1), we have 2

2



a −



1  =1 2

a=4 ∴ T(n) = aRn – 1



n −1

= 4 −



1  2

= (–1)n – 1 ⋅ 23 – n ∴ The general term is 23 – n or (–1)n – 1 ⋅ 23 – n. 22. Let a and R be the first term and the common ratio respectively. T(2) = aR = –9 ………(1) T(6) = aR5 = −

729 ………(2) 256

14

(2) ÷ (1),

R4 = R=±

n− 1 n  or ( −1) n −1 ⋅2 2 ⋅3 2  

81 256

3 4

3 into (1), we have 4

By substituting R =

3  = –9 4

a

a = –12 ∴ T(n) = aRn – 1 n −1

3   4 

= (–12) 

By substituting R = −

 

a −

3 into (1), we have 4

3  = –9 4



3  4

n −1

3   4 

∴ The general term is (–12) 

 

(or –24–2n ⋅ 3n) or

n −1

12 −

3  4

25. Let A be the surface area of the pond and a be the area covered by the lotus leaves after 1 week. ∵ R=2 and after 16 weeks, area covered = A ∴ a(216 – 1) = A

(or (–1)n–1 ⋅ 24 – 2n ⋅ 3n).

23. Let a and R be the first term and the common ratio respectively. T(3) = aR2 = 6 3 ………(1) T(5) = aR4 = 36 (2) ÷ (1),

log 6 −log 5 log 1.1025 −log 1.05

n > 3.7 ∴ The number of years taken is 4.

n −1



………(2) R =6 R=± 6 3

2

By substituting R =

 .  

24. Let n be the number of years taken. Consider the salary of Leo, a = $10 000, R = 1 + 10.25% = 1.1025 ∴ The salary of Leo after n years = $10 000 × 1.1025n Consider the salary of Michael, a = $12 000, R = 1 + 5% = 1.05 ∴ The salary of Michael after n years = $12 000 × 1.05n For Leo to have a salary higher than that of Michael, we have $10 000 × 1.1025n > $12 000 × 1.05n 5 × 1.1025n > 6 × 1.05n log (5 × 1.1025n) > log (6 × 1.05n) log 5 + n log 1.1025 > log 6 + n log 1.05 n (log 1.1025 – log 1.05) > log 6 – log 5

n>

a = 12 ∴ T(n) = aRn – 1 = 12  −

Arithmetic and Geometric Sequences and their Summation

6 into (1), we have

2

a ( 6 ) =6 3 a= 3 ∴ T(n) = aRn – 1 = 3 ( 6 ) n −1

a=

A 215

Let it takes k weeks to cover one eighth of the pond. ∴ a(2k – 1) =

A 8

A A ( 2 k −1 ) = 15 8 2 2k – 1= 212 k = 13 ∴ It takes 13 weeks to cover one eighth of the pond. 26. (a) Let the vertices of the kth triangle be Ak, Bk and Ck, k = 1, 2, 3,…

By substituting R = − 6 into (1), we have a ( − 6 ) 2 =6 3 a= 3 ∴ T(n) = aRn – 1 = 3 ( − 6 ) n−1

∴ The general term is n− 1 n  or 2 2 ⋅3 2  

80

    or

3(

6 ) n−1

3 ( − 6 ) n−1

C2B2 =

1 C1B1 2

(mid-pt. theorem)

Certificate Mathematics in Action Full Solutions 5A

1 B 1A 1 2 1 A 2C 2 = A 1C 1 2 B 2A 2 =

5 000 000 × (1 + 4%)n > 2 × 5 000 000 n log 1.04 > log 2 n > 17.7 ∴ At the end of 2018, the population will be doubled.

(mid-pt. theorem) (mid-point theorem)

∴ Perimeter of second triangle = C2B2 + B2A2 + A2C2

29. Let the original numbers be k, 5k and 11k respectively. ∵ k + 3, 5k + 3, 11k + 3 is a geometric sequence.

1 1 1 C1B1 + B 1A 1 + A1C1 2 2 2 1 = × perimeter of first triangle 2 =

∴

∴

Perimeter of second traingle Perimeter of first trai ngle

=

11 k + 3 5k + 3 = 5k +3 k +3 (5k + 3)2 = (11k + 3)(k + 3) 25k2 + 30k + 9 = 11k2 + 36k + 9 14k2 – 6k = 0 k(7k – 3) = 0

1 2

3 or k = 0(rejected) 7 3 15 33 , and . ∴ The original numbers are 7 7 7 k=

(b) By an argument similar to (a), Perimeter of third triangle

1 × perimeter of second triangle 2 1 1 = × × perimeter of first triangle 2 2 =

Exercise 14E (p. 181) Level 1

2

1.

1  x 2

=

(a) Geometric mean =

9 ×27 =9

(b) Geometric mean = −

3

4× 16 =−8

4

1  x 2

(c) Geometric mean =

∴ Perimeter of fifth triangle = 

± −4 × (− 36 ) = 12 or − 12

4

i.e.

1  x 2

1.5 = 

2.

∵ The geometric mean between x and 48 is 12. ∴ 122 = 48x x=3

3.

(a) Let R be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: 243, 243R, 243R2, 9 ∵ The 4th term is also given by 243R3. ∴ 243R3 = 9

x = 24 27. Let the height of the building be h m.

3 ∵ R= 5 4

3  h 5

∴ After the 4th rebound, the height = 

4

i.e.

3 1.62 =   h 5

h = 12.5 ∴ The height of the building is 12.5 m. 28. (a) At the end of 2001, population = 5 000 000 × (1 + 4%) = 5 200 000 At the end of 2002, population = 5 200 000 × (1 + 4%) = 5 000 000 × (1 + 4%)2 = 5 408 000 (b) Let the population will be doubled at the end of the nth year.

R=

1 3

∴ The two required geometric means are 81 and 27. (b) Let r be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: –16, –16r, –16r2, –16r3, –81 ∵ The 5th term is also given by –16r4. ∴ –16r4 = –81 r=±

3 2

∴ The three required geometric means are 24, –36 and 54 or –24, –36 and –54.

14

∴ n is the arithmetic mean between –2 and m. ∴ 2n = m + (–2) m = 2n + 2 ……(2) By substituting (2) into (1), we have (2n + 2)n = 4 n2 + n – 2 = 0 (n – 1)(n + 2) = 0 n =1 or n = –2 (rejected) By substituting n = 1 into (2), we have m = 2(1) + 2 = 4

(c) Let r be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: 1, r, r2, r3, r4, 32 ∵ The 6th term is also given by r5. ∴ r5 = 32 r=2 ∴ The four required geometric means are 2, 4, 8 and 16. 4.

Let R be the common ratio. The geometric sequence is: 54, 54R, 54R2, 2 ∵ The 4th term is also given by 54R3. ∴ 54R3 = 2 R=

Arithmetic and Geometric Sequences and their Summation

9.

1 3

∴ The value of a and b are 18 and 6 respectively.

(a) b = a + a tan θ = a(1 + tan θ ) c = b + b tan θ = b(1 + tan θ ) ac = a[b(1 + tan θ )] = b[a(1 + tan θ )] = b2 ∴ b is a geometric mean between a and c. (b) ∵

5. 2, 18 or 3, 12 or –2, –18 or –3, –12 (or any other reasonable answers)

∴ The common ratio is 1 + tan θ . If θ = 30°, 1 + tan θ = 1 + tan 30°

Level 2 6.

∵ x + 1, x + 5, 2x + 4 are in geometric sequence. ∴ x + 5 is the geometric mean between x + 1 and 2x + 4. ∴ (x + 5)2 = (x +1)(2x + 4) 2 x + 10x + 25 = 2x2 + 6x + 4 x2 – 4x – 21 = 0 (x + 3) (x – 7) = 0 3 or 7 x= −

b a(1 + tan θ ) = = 1 + tan θ a a

=

1.

(a) ∵ a = 1, d = 3 – 1 = 2 and n = 20

(b) ∵ a = –5, d = –2 – (–5) = 3 and n = 25 ∴ S(25) =

a2 3 . R 2

a2 3 4 R = 2 a 2 R= a

(c) ∵ a = 78, d = 72 – 78 = –6 and n = 27 ∴ S(27) =

(d) ∵ first term = (a + b), d = (3a – b) – (a + b) = 2a – 2b and n = 10

(3 2 ) 2 = 2a

∴ S(10) =

a=9 ∵ m, –2, n are in geometric sequence. ∴ –2 is the geometric mean between m and n. ∴ (–2)2 = mn mn = 4 ……(1) ∵ –2, n, m are in arithmetic sequence.

80

27 [2(78) + (27 – 1)(–6)] 2

=0

(b) ∵ The geometric mean between a and 2 is 3 2 .

8.

25 [2(–5) + (25 – 1)(3)] 2

= 775

∴ The two geometric means are a and 2.

∴

20 [2(1) + (20 – 1)(2)] 2

= 400

a2 a2 a2 2 4 , R, R , 2 2 2 a

∴

+1

Level 1

(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

3

Exercise 14F (p. 189)

∴ S(20) = 7.

1

10 [2(a + b) + (10 – 1)(2a – 2b)] 2

= 100 a −80 b 2.

(a) Let a and d be the first term and the common difference respectively. ∵ a = 1 and d = 9 – 1 = 8 ∴ T(n) = a + (n – 1)d

Certificate Mathematics in Action Full Solutions 5A

= 1 + (n – 1)(8) = 8n – 7 Let 97 be the kth term. i.e. T(k) = 97 ∴ 8k – 7 = 97 k = 13 ∴ There are 13 terms in the sequence. ∴ S(13) =

∵ a=

∴ T(n) = a + (n – 1)d

1 1 + ( n −1) 3 6 1 1 = + n 6 6 =

13 (1 + 97 ) 2

5 be the kth term. 6 5 i.e. T(k) = 1 6 1 1 5 + k =1 ∴ 6 6 6

= 637

Let 1

(b) Let a and d be the first term and the common difference respectively. ∵ a = –1 and d = 2 – (–1) = 3 ∴ T(n) = a + (n – 1)d = –1 + (n – 1)(3) = 3n – 4 Let 95 be the kth term. i.e. T(k) = 95 ∴ 3k – 4 = 95 k = 33 ∴ There are 33 terms in the sequence. ∴ S(33) =

k = 10 ∴ There are 10 terms in the sequence. ∴

33 (−1 + 95 ) 2

5 1 10  +1  6 S(10) = 3 2 5

= 10 6

= 1551 (c) Let a and d be the first term and the common difference respectively. ∵ a=5

3 1 3 3 and d = 6 − 5 = 4 2 4 4

∴ T(n) = a + (n – 1)d

3 3 + ( n −1) 4 4 4 =5 + n 3 =5

3.

(a) Let N be the number of terms that must be taken. ∵ a = 7, d = 10 – 7 = 3 and S(n) = 920

 3  36 5 + 32  S(36) =  4  2 =679

S(n) =

∴

1840 = 3N2 – 11N 3N – 11N – 1840 = 0 (N – 23)(3N + 80) = 0 2

N = 23 or −

80 (rejected) 3

∴ 23 terms of arithmetic series must be taken.

3 k = 32 4

k = 36 ∴ There are 36 terms in the sequence. ∴

N [2a + (N – 1)d] 2 N 920 = [2(7) + (N – 1)(3)] 2

and

Let 32 be the kth term. i.e. T(k) = 32 ∴ 5+

1 1 1 1 and d = − = 3 2 3 6

1 2

(d) Let a and d be the first term and the common difference respectively.

(b) Let N be the number of terms that must be taken. ∵ a = 21, d = 15 – 21 = –6 and S(n) = –60 and ∴

N [2a + (N – 1)d] 2 N –60 = [2(21) + (N – 1)(–6)] 2 S(n) =

–120 = –6N2 + 48N N2 – 8N – 20 = 0 (N – 10)(N + 2) = 0 N = 10 or –2(rejected) ∴ 10 terms of arithmetic series must be taken. (c) Let N be the number of terms that must be taken. ∵ a = 27, d = 24 – 27 = –3 and S(n) = 126

14

N [2a + (N – 1)d] 2 N 126 = [2(27) + (N – 1)(–3)] 2

and

8 [2a + (8 – 1)(–4)] = 336 2

S(n) =

∴

252 = –3N2 + 57N N2 – 19N + 84 = 0 (N – 7)(N – 12) = 0 N = 7 or 12 ∴ 7 or 12 terms of arithmetic series must be taken.

a = 56 ∴ The first term is 56.

7.

8.

(b) Let n be the number of terms of the given series. ∵ a = 6, d = 12 – 6 = 6 and l = T(n) = 198 and T(n) = a + (n – 1)d ∴ 198 = 6 + (n – 1)(6) n = 33 ∴ The required sum =

(c) The required sum = sum of integers between 1 and 200 inclusive – sum of integers between 1 and 200 that are multiples of 6 = 20 100 – 3366 (from (a) and (b)) = 16 734

S(14) = 406

14 [2(a) + (14 – 1)d] = 406 2

2a +13d = 58 ………(1) ∵ T(4) + T(5) = 34 ∴ (a + 3d) + (a + 4d) = 34 2a +7d = 34 ………(2) (1) – (2), 6d = 24 d=4 By substituting d = 4 into (2), we have 2a + 7(4) = 34 a=3 ∴ The first term is 3 and the common difference is 4. (b) S(20) =

20 [2(3) + (20 – 1)(4)] 2

= 820 6.

∵ d = –4 and

33 (6 +198 ) 2

= 3366

= 130

∴

200 (1 + 200 ) 2

= 20100

∵ d = 2 and T(4) = 10 ∴ a + 3(2) = 10 a=4

(a) ∵

(a) ∵ a = 1, l = 200 and n = 200 ∴ 1 + 2 + … + 200 =

10 ∴ S(10) = [2(4) + (10 – 1)2] 2

5.

7 [2(100) + (7 – 1)(–10)] 2

= $490 > $480 ∴ She has enough money to buy the watch.

14 820 = 4N2 + 32N N2 + 8N – 3705 = 0 (N – 57)(N + 65) = 0 N = 57 or –65(rejected) ∴ 57 terms of arithmetic series must be taken. 4.

∵ a = 100, d = –10 ∴ S(7) = $

(d) Let N be the number of terms that must be taken.

1 4 ∵ a = 6, d = 7 − 6 = and S(n) = 2470 3 3 N and S(n) = [2a + (N – 1)d] 2 4 N ∴ 2470 = [2(6) + (N – 1)   ] 2 3

Arithmetic and Geometric Sequences and their Summation

S(8) = 336

9.

(a) For a common difference of 6, the arithmetic sequence is 2, 8, 14, …, 98. Let 98 be the kth term. i.e. T(k) = 98 2 + (k – 1)(6) = 98 k = 17 ∴ 2 + 8 + 14 + … + 98 = S(17) =

17 (2 + 98 ) 2

= 850 ∴ The sum of the corresponding series is 850. For a common difference of 16, the arithmetic sequence is 2, 18, 34, …, 98. Let 98 be the jth term. i.e. T(j) = 98 2 + (j – 1)(16) = 98 j=7 ∴ 2 + 18 + 34 + … + 98 = S(7) =

7( 2 + 98 ) 2

= 350

80

100

Certificate Mathematics in Action Full Solutions 5A

∴ The sum of the corresponding series is 350. (or any other reasonable answers) (b) For a common difference of 8, the arithmetic sequence is 2, 10, 18, …, 98. Let 98 be the kth term. i.e. T(k) = 98 2 + (k – 1)(8) = 98 k = 13 ∴ 2 + 10 + 18 + … + 98 = S(13) =

13 (2 + 98 ) 2

= 650 ∴ The sum of the corresponding series is 650. For a common difference of 4, the arithmetic sequence is 2, 6, 10, …, 98. Let 98 be the kth term. i.e. T(k) = 98 2 + (k – 1)(4) = 98 k = 25 ∴ 2 + 6 + 10 + … + 98 = S(25) =

25 (2 + 98 ) 2

= 1250 ∴ The sum of the corresponding series is 1250. (or any other reasonable answers) 10. (a) (–2) + (–1) + 0 + 1 + 2 or (–6) + (–3) + 0 + 3 + 6 (or any other reasonable answers) (b) 1 + 2 + 3 + 4 + 5 or (–1) + 1 + 3 + 5 + 7 (or any other reasonable answers)

Level 2 11. (a) Let k be the number of rows. ∵ a = 12, d = 3 and T(k) = 6a ∴ a + (k – 1)d = 6a 12 + (k – 1)(3) = 6(12) k = 21 ∴ The number of rows is 21.

n [2a + (n – 1)d] 2 21 S(21) = [2(12) + (21 – 1)(3)] 2

(b) S(n) =

= 882 ∴ The number of seats is 882. 12. Let m1, m2, m3, …, m25 be the arithmetic means. ∵ a = 5, l = 120 and n = 27 ∴

S(27) =

27 (5 +120 ) 2

i.e. 5 + m1 + m2 + m3 + … + m25 + 120 = 1687.5

101

m1 + m2 + m3 + … + m25 = 1562.5 ∴ The sum of the 25 arithmetic means between 5 and 120 is 1562.5. 13. Let m1, m2, m3, …, mx be the arithmetic means. ∵ a = x, l = 2x and n = x + 2 ∴ S(x + 2) =

( x + 2)( x + 2 x ) 2 i.e. x + m1 + m2 + m3 + … + mx + 2x =

( x + 2)( x + 2 x ) 2 m1 + m2 + m3 + … + mx =

3x 2 2

∴ The sum of the x arithmetic means between x and 2x is

3x 2 . 2

14. (a) first term = T(1) = 55 – 2(1) = 53 Common difference = T(n + 1) – T(n) = [55 – 2(n + 1)] – (55 – 2n) 2 =− (b) Let the kth term be the first negative term. i.e. T(k) < 0 ∴ 55 – 2k < 0 k>

55 2

∴ The first negative term is the 28th term. ∴ T(28) = 55 – 2(28) 1 =− (c) ∵ T(n) is negative for n ≥ 28. ∴ S(n) is maximum at n = 27. S(27) =

27 [2(53) + (27 – 1)( –2)] 2

= 729 ∴ The maximum value of S(n) is 729. 15. ∵ a = –49, d = (–46) – (–49) = 3 ∴ T(n) = –49 + (n – 1)(3) = 3n – 52 Let the kth term be the last negative term. ∴ T(k) < 0 i.e. 3k – 52 < 0 k<

52 3

∴ The last negative term is the 17th term. S(17) =

17 [2(–49) + (17 – 1)(3)] 2

= –425 ∴ The sum of all negative terms is –425.

14

16. (a) The sum = 200 + 205 + 210 + … + 500 ∴ a = 200, d = 5 and l = 500 T(n) = 200 + (n – 1)(5) = 195 + 5n Let k be the number of terms. ∴ T(k) = 500 i.e. 195 + 5k = 500 k = 61

Arithmetic and Geometric Sequences and their Summation

(b) The total number of terms = 1 + 2 + 3 + … + n n(1 +n) = 2 (c) The last term in the first n brackets = the total number of terms in the first n brackets

61(200 + 500 ) S(61) = 2

=

= 21 350 ∴ The required sum is 21 350.

Now, a = 1 and l =

(b) The sum = 203 + 210 + 217 + … + 497 ∴ a = 203, d = 7 and l = 497 T(n) = 203 + (n – 1)(7) = 196 + 7n Let k be the number of terms. ∴ T(k) = 497 i.e. 196 + 7k = 497 k = 43 S(43) =

n (1 + n )   1 +  2   = 2 n(1 + n)  n  = 1 + 2 (1 + n)  4   n (1 + n ) 2

43 ( 203 + 497 ) 2

(d) The last term in the first (n – 1) brackets = The total number of terms in the first (n – 1) brackets

(c) If the integers are divisible by both 5 and 7, then they are divisible by 35. We are going to find the sum of integers divisible by 35 between 200 and 500 inclusive. ∴ The sum = 210 + 245 + 280 + … + 490 ∴ a = 210, d = 35 and l = 490 T(n) = 210 + (n – 1)(35) = 175 + 35n Let k be the number of terms. ∴ T(k) = 490 i.e. 175 + 35k = 490 k=9

( n −1)[1 + ( n −1)] 2 n( n −1) = 2 n( n −1) Now, a = 1 and l = 2 =

∴ The sum of the terms in the first (n – 1) brackets

9( 210 + 490 ) 2

= 3150 ∴ The required sum is 3150.

=2

2080

18. (a) In the 1st bracket, there is 1 term. In the 2nd bracket, there are 2 terms. In the 3rd bracket, there are 3 terms. ∴ Number of terms in the nth bracket is n.

80

(n −1)[1 + (n −1)]  ( n −1)  [1 + ( n −1)]  1 + 4 2  

=

n(n −1)  n  1 + ( n −1)  4 2  

n(1 + n)  n n  n( n −1)  1 + (1 + n)  − 1 + ( n −1) 4 2 4 2    n = (1 + n 2 ) 2 =

17. 2 × 22 × 23 × …× 264 = 21 + 2 + 3 + …. + 64 64 (1+64 ) 2

=

∴ The sum of the terms in the nth bracket = The sum of the terms in the first n brackets – the sum of the terms in the first (n – 1) brackets

(d) The required sum = sum of integers divisible by 5 + sum of integers divisible by 7 – sum of integers divisible by both 5 and 7 = 21 350 + 15 050 – 3150 = 33 250

=2

n(1 + n) . 2

∴ The sum of the terms in the first n brackets

= 15 050 ∴ The required sum is 15 050.

S(9) =

n(1 + n) 2

Exercise 14G (p. 196) Level 1 1.

(a) ∵ a = 1, R =

4 = 2 and n = 10 2

102

Certificate Mathematics in Action Full Solutions 5A

∴ S(10) =

1( 210 −1) 2 −1

∵ a=

=1023

1 1 , R =  1  = 3 and T(n) = aRn – 1 = 729 3  3  

1 (3)n – 1 3

∴ 729 = (b) ∵ a = 27, R =

∴ S(7) =

9 1 = and n = 7 27 3

n=8 ∴ The number of terms is 8.

  1 7  27 1 −      3   1−

=

1 8 (3 −1) 3 3 −1

∴ S(8) =

1

=

3

1093 27

3280 3

(d) Let n be the number of terms of the given series. 2

(c) ∵ a = 8, R =

8 1 2 , R = 51 = 2 and T(n) = aRn – 1 = 134 4 5 5 5

∵ a=4

−16 = –2 and n = 10 8

8[1 −( −2)10 ] ∴ S(10) = 1 −( −2)

∴

2 1 = 4 ( 2) n −1 5 5

134

n=6 ∴ The number of terms is 6.

=−2728

0.6 (d) ∵ a = –1, R = = –0.6 and n = 7 −1 −1[1 −( −0.6) 7 ] ∴ S(7) = 1 −( −0.6)

∴ S(6) =

4

1 6 (2 −1) 5 2 −1

= 264

3 5

=−0.642 496

3. 2.

(a) Let n be the number of terms of the given series. ∵ a = 2, R =

6 = 3 and T(n) = aRn – 1 = 4374 2

∴ 4374 = 2(3n – 1) n=8 ∴ The number of terms is 8.

2(3 −1) 3 −1 8

∴ S(8) =

=6560

(b) Let N be the number of terms of the given series.

−8 ∵ a = 2, R = = –4 and T(N) = aRN – 1 = –2048 2 ∴ –2048 = 2(–4)N – 1 N=6 ∴ The number of terms is 6. ∴ S(6) =

2[1 −( −4) 6 ] 1 −( −4)

=−1638

(c) Let n be the number of terms of the given series.

103

(a) Let N be the number of terms that must be taken.

2 a ( R N −1) = 511 = 2 and S(N) = 1 R −1 N 1( 2 −1) ∴ 511 = 2 −1 ∵ a = 1, R =

N=9 ∴ 9 terms of geometric series must be taken.

(b) Let N be the number of terms that must be taken. 1 

3  1   ∵ a= , R =  1  = 3 and 9 9   

S(N) =

a ( R N −1) 4 = 40 R −1 9 1

∴

4 40 = 9

9

(3

N

−1)

3 −1

N=6 ∴ 6 terms of geometric series must be taken. (c) Let N be the number of terms that must be taken.

14

∵ a = 48, R = and S(N) =

24 1 = 48 2

R6 + 7R3 – 8 = 0 (R + 8)(R3 – 1) = 0 R3 = –8 or R3 = 1 R = –2 or R = 1(rejected) By substituting R = –2 into (1), we have 3

  1 N  481 −     2  29  ∴ 95 =  1 32 1−

a[( −2) 3 −1] = 12 ( −2) −1

2

a=4 ∴ The first term is 4 and the common ratio is –2.

N = 10 ∴ 10 terms of geometric series must be taken. (d) Let N be the number of terms that must be taken.

6.

2 

Let n be the least number of terms taken.



N

a ( R −1) 26 = 242 R −1 27 2

∴

242

26 = 27

27

(3

N

3 2

 

7.

2

2  = 27 3 a=

243 4

243   2  1 −  −  4   3 

10

∴ S(10) =

=

 2 1− −   3

5000 (1.06 )(1.06 10 −1)   1.06 −1    

  

= 69 858 (cor. to the nearest dollar) ∴ Peter will receive $69 858 at the end of the 10th year.

11 605 324

a ( R 3 −1) = 12 R −1

Level 2 8.

……(1)

∵ T(4) + T(5) + T(6) = –96 ∴ S(6) – S(3) = –96

a ( R 6 −1) = –84 R −1

At the end of the 1st year, Peter will receive $5000(1 + 6%)1 = $5000(1.06)1. At the end of the 2nd year, he will receive $5000(1.06) + $5000(1.06)2. At the end of the 3rd year, he will receive $5000(1.06) + $5000(1.06)2 + $5000(1.06)3. ∴ a = 5000(1.06), R = 1.06 S(10) = 

Let the first term be a and the common ratio be R. ∵ S(3) = 12 ∴

−1

N > 11.38 ∴ At least 12 terms of geometric series must be taken.

3 −1

∵ T(3) = 27 ∴ a −

5.

∴

−1)

N=8 ∴ 8 terms of geometric series must be taken. 4.

6 3 a ( R N −1) > 800 = and S(N) = 4 2 R −1  3  N  4   − 1  2   > 800

∵ a = 4, R =

9  2   ∵ a= ,R= 2 =3 27  27 

and S(N) =

R 6 −1 = –7 R 3 −1

(2) ÷ (1),

a (1 − R N ) 29 = 95 1−R 32



Arithmetic and Geometric Sequences and their Summation

……(2)

At the end of the 1st year, he will get $1000(1.04). At the end of the 2nd year, he will get $1000(1.04) + $1000(1.04)2. At the end of the 3rd year, he will get $1000(1.04) + $1000(1.04)2 + $1000(1.04)3. ∴ a = 1000(1.04), R = 1.04 Let n be the number of years needed. S(n) > 15 000

1000 (1.04 )(1.04 n −1)    > 15 000 1.04 −1     n > 11.6

80

104

Certificate Mathematics in Action Full Solutions 5A

∴ The minimum number of years needed is 12. ∴ S(6) = 9.

(a) Let N be the number of terms that must be taken. ∵ a= − and S(N) = ∴

60

1 1 , R =  − 1  = –3  3 3  

a (1 − R N ) 2 = 60 1−R 3

2 = 3

−

1 3

and S(N) =

−18 1 =− 36 2

a (1 − R N ) 125 = 23 1−R 128

10. (a) (i) In the 2nd second, the distance travels = 20 × 0.8 m In the 3rd second, the distance travels = 20 × 0.82 m In the 4th second, the distance travels = 20 × 0.83 m ∴ In the nth second, the distance travels is 20 × 0.8n – 1 m. (ii) ∵ a = 20, R = 0.8

20 (1 −0.8 n ) ∴ S(n) = = 100(1 – 0.8n) 1 −0.8 ∴ The total distance travels in the first n seconds is 100(1 – 0.8n) m. (b) From (a), the distance travels in the first 18 seconds = S(18) = 100(1 – 0.818) m = 98.2 m < 100 m ∴ The train cannot stop at the station successfully in 18 seconds. 11. (a) After the 1st blow, the length driven is 2 m. After the 2nd blow, the length driven is 2(0.9) m. … After the 6th blow, the length driven is 2(0.9)5 m. ∴ a = 2 and R = 0.9

2(1 −0.9 7 ) m 1 −0.9

= 10.43 m (cor. to 2 d.p.) > 10 m ∴ The pile would be completely driven into the ground.

1 −( −3)

  1 N  36 1 −−     2   ∴ 23 125 =  128  1 1 − −   2 N = 10 ∴ 10 terms of geometric series must be taken.

105

(b) S(7) =

N

(b) Let N be the number of terms that must be taken. ∵ a = 36, R =

∴ The length of the pile driven is 9.37 m.

[1 −(−3) ]

N=6 ∴ 6 terms of geometric series must be taken.

2(1 −0.9 6 ) = 9.37, cor. to 2 d.p. 1 −0.9

12. (a)

BD = ABsin θ ∴ d1 = x sin θ AD = ABcos θ = xcos θ DE = ADsin θ = xcos θ sin θ ∴ d2 = x sin θcos θ

(b) (i) AE = ADcos θ = x(cos θ )2 d3 = EF = AEsin θ = xsin θ (cos θ )2 AF = AEcos θ = x(cos θ )3 d4 = FG = AFsin θ = xsin θ (cos θ )3

d 2 x sin θ cos θ = = cosθ d1 x sin θ

d3 x sin θ (cos θ ) 2 = = cosθ d2 x sin θ cos θ d4 x sin θ (cos θ ) 3 = = cosθ d3 x sin θ (cos θ ) 2 ∴

d 2 d3 d 4 = = d1 d 2 d 3

∴ d1, d2, d3, d4 are in geometric sequence. (ii) ∵ a = xcos θ , R = cos θ ∴ d1 + d2 + d3 + d4 = S(4)

x sin θ[1 −cos 4 θ] 1 −cos θ = x sin θ(1 +cos θ)(1 +cos =

(c) d1 + d2 + d3 + d4 = 20sin 30°(1 + cos 30°)(1 + cos230°) 35 (2 + 3 ) = 4 13. (a) 99 – 9 = 90 999 – 99 = 900 ≠ 90 ∴ It is not an arithmetic sequence.

2

θ)

14

Arithmetic and Geometric Sequences and their Summation

=1+2+3 ∴ The number of terms in the first n brackets = 1 + 2 + 3 +… + n

99 = 11 9 999 111 = ≠ 11 99 11

=

∴ It is not a geometric sequence.

n2 + n 2

and a = 1, R = 2

1

(b) 9 = 10 – 1 = 10 – 1 99 = 100 – 1 = 102 – 1 999 = 1000 – 1 = 103 – 1 ∴ T(n) = 10n – 1

∴ The sum =

=2

(c) T(1) + T(2) + T(3) + … + T(n) = (101 – 1) + (102 – 1) + (103 – 1) + … + (10n – 1) = 101 + 102 + 103 + … + 10n – n

10 (10 n −1) = −n 10 −1 10 = (10 n −1) −n 9

=

1 × first term in the (n + 1)th bracket 2

=

( n +1)( n +1−1) 1 2 ×2 2

(b) The number of terms in the 1st bracket = 1 The number of terms in the 2nd brackets = 2 The number of terms in the 3rd brackets = 3 ∴ The number of terms in the nth brackets = n

∴ The sum =

2

( 2 −1) 2 −1 n

= ( 2 n −1)2

n ( n −1) 2

(c) The number of terms in the first bracket = 1 The number of terms in the first 2 brackets =1+2 The number of terms in the first 3 brackets

80

2 2

2 4

2 2(n – 1)

= ( r +1)

a 2 (r 2 n −1) r 2 −1

= (r + 1)S2

Exercise 14H (p. 203)

,R=2

n ( n −1) 2

2

a 2 [( r 2 ) n −1] r 2 −1 a 2 ( r 2 n −1) = r 2 −1 =

2

n 2 +n −2 2

n ( n −1) 2

a ( r n −1) r −1

 a (r n −1)  a (r n −1)  = ( r −1) + 2 a  r −1  r −1   2 n 2 2 n a ( r −1) 2a ( r −1) = + r −1 r −1 2 n a ( r −1) n = (r +1) r −1 a 2 ( r 2 n −1) = r −1 2 a ( r 2 n −1) (r +1) = ⋅ r −1 (r +1)

n ( n −1) 2

a =2

−1

(r – 1)S12 + 2aS1

Last term in the nth bracket

and

n 2 +n 2

S =a +ar +ar +…+ar

14. (a) First term in the 1st bracket = 1= 20 First term in the 2nd bracket = 2 = 20 + 1 First term in the 3rd bracket = 23 = 20 + 1 + 2 First term in the 4th bracket = 26 = 20 + 1 + 2 + 3 ∴ First term in the nth bracket = 20 + 1 + 2 + … + (n – 1 )

= 2

−1) 2 −1

15. S1 = a + ar + ar2 + … + arn – 1 =

2

=2

1( 2

n 2 +n 2

Level 1 1.

(a) ∵ a = 9 and R =

6 2 = 9 3

9 ∴ S(∞) = 1 − 2 =27

3

106

Certificate Mathematics in Action Full Solutions 5A

(b) ∵ a = –4, R =

∴

= 0.747 + 0.000 747 + 0.000 000 747 + …

2 1 =− −4 2

0.747 1 −0.001 83 = 111 =

−4 1 S(∞) = 1 −  −   2 8 =− 3

3.

∴

0.2 = 0.2 1 1 ∴ S(∞) = 1 −0.2

(c) ∵ a = 1, R =

=

5 4

− 5   3  = −1 5 3

(d) ∵ a = 5, R = 

∴

5 S (∞) =  1 1 − −   3

40 7 10 40 = 1− R 7 3 R =− 4

S (∞) =

15 45 and . 2 8

(c) ∵

 = 0.7777… (a) 0.7 = 0.7 + 0.07 + 0.007 + 0.0007 + …

∴

0.7 = 1 −0.1 7 = 9

S (∞) = 90

a = 90 1 −0.2

a = 72 ∴ The first 3 terms are 72, 14.4 and 2.88. (d) ∵

S (∞) = −5

∴

7  = 0.474 747… (b) 0.4 = 0.47 + 0.0047 + 0.000 047 + … 0.47 1 −0.01 47 = 99 =

(c)

4  = 0.234 343… 0.23 = 0.2 + 0.034 + 0.000 34 + 0.000 0034 + …

0.034 = 0 .2 + 1 −0.01 116 = 495

 47  = 0.747 747… (d) 0.7 107

6 18 and . 5 25

∴ The first 3 terms are 10, −

15 = 4 2.

2 =5 1 −R 3 R= 5

∴ The first 3 terms are 2,

(b) ∵

∴

S (∞) =5

(a) ∵

a = –5 1 −( −0.2)

a = –6 ∴ The first 3 terms are –6, 1.2 and –0.24. 4.

∵ T(2) = 6 aR = 6 ……(1) and S (∞) = 24

a = 24 1 −R (1) ÷ (2),

……(2)

R(1 – R) = 2

1 4

1  R −  = 0 2  1 R= 2

14

By substituting R =

1 into (1), we have 2

8.

Arithmetic and Geometric Sequences and their Summation

(a) a = 120°, R =

1 =6 2

∴ Total angle swings through = S (∞)

a

a = 12 ∴ The first term is 12. 5.

and

1 = 24 2

a = 12 ∴ The first 3 terms are 12, 6 and 3. (a) ∵ a = 1, R = x a 1 ∴ S (∞) = 1 −R = 1 − x

1 8 1 1 =1 1 −x 8 1 x= 9

(b) ∵ S (∞) =1

7.

=360



3

 360       360 

= 20 πcm

1 – R3 =

1−

∴

2

= (2π )(10 cm)  

……(2)

a

6.

1−

(b) ∵ Total angle swings through is 360°. ∴ Total distance swings through

7 8 1 R= 2 1 By substituting R = into (2), we have 2 (1) ÷ (2),



……(1)

S (∞) = 24

a = 24 1 −R

120

=

∵ T(1) + T(2) + T(3) = 21

a (1 − R 3 ) = 21 1−R

80  2 =  3 120

The total possible output of gold = [1000 + 1000(80%) + 1000(80%)(80%) + ….] kg = [1000 + 1000(0.8) + 1000(0.8)(0.8) + ….] kg 1000 = kg 1 −0.8 =5000 kg

9.

For the downwards distance travelled, a = 10, R = 75% = 0.75 ∴

a m 1−R 10 = m 1 − 0.75

S (∞) =

= 40 m For the upwards distance travelled, a = 10(75%) = 7.5, R = 0.75 ∴

S (∞) =

a 7.5 m= m = 30 m 1 −R 1 − 0.75

∴ The total distance travelled = (30 + 40) m = 70 m 10. (a) (i) The fraction of the original piece of cake P 1 gets the first time = 4 (ii) The fraction of the original piece of cake P

1 1 1  =  4  4 16

gets the second time alone = 

(iii) The fraction of the original piece of cake P gets the nth time alone n −1

1  =  4  1 = 2n 2

1 4

Level 2

80

108

Certificate Mathematics in Action Full Solutions 5A

1 (b) ∵ a = ,R= 4

∴

1  16    =1 1  4 4   

1    4 S (∞) =   1 1− 4 =

1 3

11. (a) ∵ The speed of Ken is twice that of Angel. ∴ The distance travelled by Ken is twice that of Angel in the same time.

1 1 AB = (24 m) = 12 m 2 2 1 1 Similarly, CD = BC = (12 m) = 6 m 2 2 1 1 and DE = CD = (6 m) = 3 m 2 2

∴ BC =

BC 6 1 = = 12 2 (b) AB CD 3 1 = = BC 6 2

1 . 2

1 (c) a = 24 m, R = 2 ∴ Total distance Ken must run = S (∞)

=

24 1−

1 2

= 48 m

m

∵ △ABC is an equilateral triangle. ∴ ∠ ABC = 60° ∵ Area of △ABC = 3 × Area of △OAB ∴

1 1 (AB)(BC)(sin ∠ ABC) = 3 × (AB)(OD) 2 2 1 1 (8)(8)(sin 60°) = 3 × (8)(r1) 2 2 r1=

∴ AB, BC, CD, DE are in geometric sequence with common ratio

12. (a)

4 3

Consider △OEF. ∠ FOE = 60° OE = OF cos ∠ FOE r1 – r2 = (r1 + r2)cos 60° r2 =

1 r1 3

4 r2 = 3 3

Similarly, r3 =

4 1 r2 = 9 3 3

(b) From (a), we know that r1, r2, r3, … are in geometric sequence with first term

4 3

and common ratio

Sum of the circumferences = (2π r1 + 2π r2 + 2π r3 + …) cm = 2π (r1 + r2 + r3 + …) cm  4    3   = 2π  cm 1  1−    3    = 4 3π cm

(c) Consider the sequence r12, r22, r32, …

109

1 . 3

14

r2 2 r1

2

Arithmetic and Geometric Sequences and their Summation

BC B2 C 2 3 = 3 3 = ... = B1C1 B2 C 2 4

2

r  1 =  2  = 9  r1 

∴ B1C1, B2C2, B3C3, … are in geometric sequence.

∴ r12, r22, r32, … are in geometric sequence with first term

16 1 and common ratio . 9 3

Sum of areas of these circles = (π r21 + π r22 + π r32 + …) cm2 = π (r12 + r22 + r32 + …) cm2

 16     3 =π   cm 2 1   1 − 9    = 6π cm 2

4

3 (ii) B C =   a 4 4

4

=

81 a 256

(iii) The areas of the squares are in geometric sequence with first term ratio

9 2 a and common 16

9 . 16

Sum of areas = (B1C1)2 + (B2C2)2 + (B3C3)2 +…

13. (a) C1C = B1C1 = b AC1 = AC – C1C = 3a – b By similar triangles, we have

 9 2  a =  16 9  1− 16  9 = a2 7

AC1 B1C1 = AC BC 3a − b b = 3a a 3 b= a 4

     

14. (a) A1B1 = B1C1 = 49 cm (b) (i) From (a), we have B1C1 =

3 BC 4

Similarly, we have B2C2 = =

3 B1C1 4 3 b 4

3 A1B1 = 21 cm 7 4 B1B2 = B1C1 = 28 cm 7 A2B1 =

=

AB 2

2

A2 B12 +B1 B2 2 cm

= 21 2 +28 2 cm =35 cm

(b) Common ratio =

3 b 4 33  =  a 44 

(Pyth. theorem)

A2 B2 5 = A1 B1 7

(ii) B2C2 =

9 = a 16

(c) (i) ∵ The sides of the squares are in geometric sequence. ∴ The perimeters of the squares are also in geometric sequence with the same common ratio and first term 196 cm. Sum of the perimeters = (196 + 140 + … ) cm

(c) (i) By considering triangles AB1C1, AB2C2, …and using argument similar to (a) and (b), we have

80

110

Certificate Mathematics in Action Full Solutions 5A

T(3) = 33 – 1 = 9 , T(4) = 34 – 1 = 27 ,

   196  cm = 1 − 5    7   =686 cm

T(5) = 35 – 1 = 81 (c) T(1) = (–1)1 + 1 ⋅ 2(1) = 2 , 4 , T(2) = (–1)2 + 1 ⋅ 2(2) = −

T(3) = (–1)3 + 1 ⋅ 2(3) = 6 ,

(ii) Notice that

8 T(4) = (–1)4 + 1 ⋅ 2(4) = −

4 A1B1 = 28 cm 7 4 A 2A 3 = A2B2 = 20 cm 7 A 1A 2 =

… A1A2, A2A3, …are also in geometric sequence with the same common ratio as A1B1, A2B2, … and first term 28 cm. ∴ Total distance travelled by the ant = (28 + 20 + … ) cm    28  cm = 5   1 −  7   =98 cm 2

(d) Common ratio =

=

A2 B2 2 A1 B1 35 2 49 2

25 = 49 First term = A1B12 = 492 cm2 = 2401 cm2 ∴ The sum of areas = (2401 + 1225 + …) cm2    2401  cm 2 = 25   1 −  49   117 649 = cm 2 24

Revision Exercise 14 (p. 209) Level 1 1.

(a) T(1) = 2(1) – 1 = 1 , T(2) = 2(2) – 1 = 3 , T(3) = 2(3) – 1 = 5 , T(4) = 2(4) – 1 = 7 , T(5) = 2(5) – 1 = 9 (b) T(1) = 31 – 1 = 1 , T(2) = 32 – 1 = 3 ,

T(5) = (–1)5 + 1 ⋅ 2(5) = 10 (d) T(1) = (1)1 + 1 = 1 , T(2) = (2)2 + 1 = 8 , T(3) = (3)3 + 1 = 81 , T(4) = (4)4 + 1 = 1024 T(5) = (5)5 + 1 = 2.

15 625

(a) (i) 6, 7 (ii) ∵ T(1) = 2 = 1 + 1 T(2) = 3 = 2 + 1 T(3) = 4 = 3 + 1 T(4) = 5 = 4 + 1 ∴ T(n) = n +1 (b) (i) 20, 24 (ii) ∵ T(1) = 4 = 4(1) T(2) = 8 = 4(2) T(3) = 12 = 4(3) T(4) = 16 = 4(4) ∴ T(n) = 4n (c) (i) 4, 2 (ii) ∵ T(1) = 64 = 27 – 1 T(2) = 32 = 27 – 2 T(3) = 16 = 27 – 3 T(4) = 8 = 27 – 4 7 −n ∴ T(n) = 2 (d) (i)

3 3 , 32 64

3 3 = 1 2 2 3 3 = 2 T(2) = 4 2 3 3 T(3) = = 3 8 2 3 3 = 4 T(4) = 16 2

(ii) ∵ T(1) =

∴ T(n) =

3 2n

,

14

3.

(a) T(2) – T(1) = 4 – 1 = 3 T(3) – T(2) = 7 – 4 = 3 T(4) – T(3) = 10 – 7 = 3 ∴ It is an arithmetic sequence with common difference 3. T(n) = 1 + (n – 1)(3) = 3n −2

(d) T(2) – T(1) = (x + 3) – (x + 1) = 2 T(3) – T(2) = (x + 5) – (x + 3) = 2 T(4) – T(3) = (x + 7) – (x + 5) = 2 ∴ It is an arithmetic sequence with common difference 2. T(n) = (x + 1) + (n – 1)(2) = x +2n −1 4.

(a) (i) T(n) = 2 + (n – 1)(4) = 4 n −2

=

(b) (i) T(n) = –5 + (n – 1)(–4) 4n − 1 =− (ii) T(9) = –4(9) – 1 37 =−

3(9) − 7 2

= 10

80

10 − 7(9) 4

=− 5.

53 4

(a) Let a and d be the first term and the common difference respectively. ∵ a = 101 and d = 99 – 101 = –2 ∴ T(n) = 101 + (n – 1)(–2) = 103 – 2n Let –1 be the kth term. i.e. T(k) = –1 ∴ 103 – 2k = –1 k = 52 ∴ There are 52 terms in the sequence. (b) Let a and d be the first term and the common difference respectively. ∵ a = 3 and d = 1 – 3 = –2 ∴ T(n) = 3 + (n – 1)(–2) = 5 – 2n Let –15 be the kth term. i.e. T(k) = –15 ∴ 5 – 2k = –15 k = 10 ∴ There are 10 terms in the sequence. (c) Let a and d be the first term and the common difference respectively.

1 1 1 and d = 1 − = 2 2 2 1 1 ∴ T(n) = + (n – 1)   2 2 1 = n 2 Let 10 be the kth term. i.e. T(k) = 10

 1   2

(c) (i) T(n) = –2 + (n – 1) 1

(ii) T(9) =

(ii) T(9) =

10 −7 n 4

∵ a=

(ii) T(9) = 4(9) – 2 = 34

3n −7 = 2

3  7 + (n −1) −  4  4

(d) (i) T(n) =

(b) T(2) – T(1) = 7 – 9 = –2 T(3) – T(2) = 5 – 7 = –2 T(4) – T(3) = 2 – 5 = –3 ≠ –2 ∴ It is not an arithmetic sequence. (c) T(2) – T(1) = log 25 – log 5 = 2 log 5 – log 5 = log 5 T(3) – T(2) = log 125 – log 25 = 3 log 5 – 2 log 5 = log 5 T(4) – T(3) = log 625 – log 125 = 4 log 5 – 3 log 5 = log 5 ∴ It is an arithmetic sequence with common difference log 5. T(n) = log 5 + (n – 1)(log 5) = n log 5

Arithmetic and Geometric Sequences and their Summation

∴

1 k = 10 2

k = 20 ∴ There are 20 terms in the sequence. (d) Let a and d be the first term and the common difference respectively. ∵ a=

1 2 1 and d = − − = –1 3 3 3

112

Certificate Mathematics in Action Full Solutions 5A

d=7 By substituting d = 7 into (1), we have a + 3(7) = 75 a = 54 ∴ T(n) = a + (n – 1)d = 54 + (n – 1)(7) = 47 +7 n

1 + (n – 1)(–1) 3 4 = –n 3 2 Let −18 be the kth term. 3 2 i.e. T(k) = −18 3 4 2 −k = −18 ∴ 3 3 ∴ T(n) =

(d) Let a and d be the first term and the common difference respectively. T(7) = a + 6d = 62 ……(1) T(19) = a + 18d = 2 ……(2) (2) – (1), 12d = –60 d = –5 By substituting d = –5 into (1), we have a + 6(–5) = 62 a = 92 ∴ T(n) = a + (n – 1)d = 92 + (n – 1)(–5) = 97 −5n

k = 20 ∴ There are 20 terms in the sequence. 6.

(a) Let a and d be the first term and the common difference respectively. T(7) = a + 6d = 20 ……(1) T(19) = a + 18d = 56 ……(2) (2) – (1), 12d = 36 d=3 By substituting d = 3 into (1), we have a + 6(3) = 20 a=2 ∴ T(n) = a + (n – 1)d = 2 + (n – 1)(3) = 3n −1

7.

5 2 5 By substituting d = − into (1), we have 2  5 a + 9 −  = 40  2 125 a= 2 ∴ T(n) = a + (n – 1)d

125  5 +( n −1)−  2  2 5 = 65 − n 2 =

(c) Let a and d be the first term and the common difference respectively. T(4) = a + 3d = 75 ……(1) T(10) = a + 9d = 117 ……(2) (2) – (1), 6d = 42

113

− 8 + ( −2) 2

= –5 (b) Arithmetic mean =

117 + 49 2

= 83

(b) Let a and d be the first term and the common difference respectively. T(10) = a + 9d = 40 ……(1) T(16) = a + 15d = 25 ……(2) (2) – (1), 6d = –15 d=−

(a) Arithmetic mean =

(c) Arithmetic mean =

−13 +13 2

=0 8.

(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: 3, 3 + d2, 3 + 2d2, 3 + 3d2, 19 ∵ The 5th term is also given by 3 + 4d2. ∴ 3 + 4d2 = 19 d2 = 4 ∴ The three required arithmetic means are 7, 11 and 15. (c) Let d be the common difference of the arithmetic sequence. ∵ T(1) = a and T(4) = b ∴ b = a + 3d

14

d=

b −a 3

T (3) x4 y x = 3 2 = T ( 2) y x y T ( 4) x5 x = 4 = T (3) y x y

(i) x = T(2) =a+d

b −a    3 

∴ It is a geometric sequence with common

= a +

ratio

2 a +b = 3

n −1

b −a   = a + 2  3 

9.

a + 2b 3

=x n +1 ⋅ y 4−n

10. (a) (i) ∵ a = 1 and R = 2 and T(n) = aRn – 1 ∴ T(n) = 1(2)n – 1 n− 1 =2

T (2) −2 1 = = (a) T (1) −4 2 T (3) −1 1 = = T ( 2) − 2 2

(ii) T(10) =

210 −1

9 = 512 (or 2 )

− 1   2 1 T ( 4) = = T (3) −1 2

(b) (i) ∵ a = 2 and R = –3 and T(n) = aRn – 1 1 3) n − ∴ T(n) = 2 ⋅( −

∴ It is a geometric sequence with common

1 ratio . 2

(ii) T(10) = 2(–3)10 – 1 =

n −1

1  ∴ T(n) = − 4  2  =−2 3−n

T ( 2) 5 = (b) T (1) 2 T (3) 11 5 = ≠ T ( 2) 5 2 ∴ It is not a geometric sequence. (c)

x . y

2 3 x  ∴ T(n) = x y  y   

(ii) y = T(3) = a + 2d

=

Arithmetic and Geometric Sequences and their Summation

T ( 2) 0.33 11 = = T (1) 0.3 10 T (3) 0.333 111 11 = = ≠ T ( 2) 0.33 110 10 ∴ It is not a geometric sequence.

T ( 2) x3 y 2 x = 2 3 = (d) T (1) y x y

− 39 366 (or −2 ⋅39 )

(c) (i) ∵ a = 1 and R = –1 and T(n) = aRn – 1 ∴ T(n) = 1(–1) n – 1 1 1) n − = (− (ii) T(10) = (–1)10 – 1 1 =−

1 and R = –2 and T(n) = aRn – 1 4 1 n −1 ∴ T(n) = − ( −2) 4

(d) (i) ∵ a = −

=( −1) n ⋅2 n −3

(ii) T(10) = ( −1)10 ( 210 −3 ) = 128

(or 2 7 )

11. (a) Let a and R be the first term and the common ratio respectively. ∵ a = 1 and R =

2 =2 1

∴ T(n) = aRn – 1

80

114

Certificate Mathematics in Action Full Solutions 5A

= 1(2)n – 1 = 2n – 1 Let 2048 be the kth term. i.e. T(k) = 2048 2k – 1 = 2048 k = 12 ∴ The number of terms is 12. (b) Let a and R be the first term and the common ratio respectively. ∵ a = –2 and R =

1 9 1 a= 27

a(3) =

∴ T(n) = aRn – 1 =

−10 =5 −2

∴ T(n) = aRn – 1 = –2(5)n – 1 Let –6250 be the kth term. i.e. T(k) = –6250 –2(5)k – 1 = –6250 k=6 ∴ The number of terms is 6. (c) Let a and R be the first term and the common ratio respectively. ∵ a = 3.2 and R =

R=3 By substituting R = 3 into (1), we have

1.6 1 = 3.2 2

∴ T(n) = aRn – 1 n −1

1   2

= 3.2 

=3

T(5) = aR4 =

−12 ∵ a = 3 and R = = –4 3 ∴ T(n) = aRn – 1 = 3(–4)n – 1 Let 12 288 be the kth term. i.e. T(k) = 12 288 3(–4)k – 1 = 12 288 k=7 ∴ The number of terms is 7. 12. (a) Let a and R be the first term and the common ratio respectively.

6

………(1)

T(7) = aR = 27 ………(2) (2) ÷ (1), R5 = 243

115

R=

3

1  =2 4

a

a = 128 ∴ T(n) = aRn – 1

1   4 

= 128  =2

= 0.006 25

(d) Let a and R be the first term and the common ratio respectively.

1 9

………(2)

n −1

k = 10 ∴ The number of terms is 10.

T(2) = aR =

1 2

1 4 1 By substituting R = into (1), we have 4 (2) ÷ (1),

k −1

1   2 

n− 4

(b) Let a and R be the first term and the common ratio respectively. T(4) = aR3 = 2 ………(1)

Let 0.006 25 be the kth term. i.e. T(k) = 0.006 25 3.2 

1 ⋅ 3n – 1 27

9−2 n

(c) Let a and R be the first term and the common ratio respectively. T(3) = aR2 = –8 ………(1) T(6) = aR5 = 1 ………(2)

1 8 1 R=− 2 1 By substituting R = − into (1), we have 2 R3 = −

(2) ÷ (1),



2

a −



1  = –8 2

a = –32 ∴ T(n) = aRn – 1



= –32  −



n −1

1  2

1) n ⋅26−n = (−

(d) Let a and R be the first term and the common ratio

14

respectively. T(2) = aR = 6 ………(1) T(7) = aR6 = 192 ………(2) (2) ÷ (1), R5 = 32 R=2 By substituting R = 2 into (1), we have a(2) = 6 a=3 ∴ T(n) = aRn – 1 n− 1 = 3 ⋅2 13. (a) Geometric mean =

∴ S(12) =

(d) ∵ first term = (a – b), d = (3a + b) – (a – b) = 2a + 2b and n = 10 ∴ S(10) =

(c) Let R be the common ratio of the geometric sequence to be formed. The geometric sequence formed is:

n( 200 + 2) = 10 100 2

n = 100 ∴ There are 100 terms in the given series. (ii) Let d be the common difference. ∵ l = T(100) = 2 ∴ 200 + 99d = 2 d = –2 ∴ The common difference is –2.

1 3 R. 4

1 3 27 R = 4 32 3 R= 2

∴ The two required geometric means are

15. (a) (i) Let n be the number of terms of the given series. ∵ a = 200, l = 2 and S(n) = 10 100 ∴

1 1 1 27 , R, R 2 , 4 4 4 32

∴

3 and 8

9 . 16

(b) (i) Let a be the first term and d be the common difference. ∵ S(9) = 90 ∴

9 (2a + 8d) = 90 2

a + 4d = 10 ………(1) T(11) = 14 a + 10d = 14 ………(2) (2) – (1), 6d = 4 ∵

(d) Let r be the common ratio of the geometric sequence to be formed. The geometric sequence formed is: 3, 3r, 3r2, 24 ∵ The 4th term is also given by 3r3. ∴ 3r3 = 24 r=2 ∴ a = 6 and b = 12

d=

20 [2(–3) + (20 – 1)(5)] 2

= 890

(ii) By substituting d = a=

15 [2(65) + (15 – 1)(–3)] 2

= 660

2 . 3

2 into (1), we have 3

22 3

∴ S(50) =

(b) ∵ a = 65, d = 62 – 65 = –3 and n = 15 ∴ S(15) =

2 3

∴ The common difference is

14. (a) ∵ a = –3, d = 2 – (–3) = 5 and n = 20 ∴ S(20) =

10 [2(a – b) + (10 – 1)(2a + 2b)] 2

= 100 a +80 b

−3 × (− 27 ) =9 or −9

∵ The 4th term is also given by

12 [2(7) + (12 – 1)( –3)] 2

114 =−

4 ×64 =16

(b) Geometric mean = ±

Arithmetic and Geometric Sequences and their Summation

50   22   2  2  + 49   2   3   3 

=

3550 3

16. (a) ∵ a = 1, l = 50

(c) ∵ a = 7, d = 4 – 7 = –3 and n = 12

80

116

Certificate Mathematics in Action Full Solutions 5A

∴ S(50) =

50 [1 + 50] 2

∴

= 1275 (b) The sum of even integers = 2 + 4 + 6 +…. + 100 = 2( 1 + 2 + 3 + … + 50) = 2(1275) (from (a)) = 2550

3N – 1 > 4000 N > 7.55 ∴ 8 terms of geometric series must be taken. (b) Let n be the number of terms that must be taken. 8 

3  2   ∵ a= , R = 2  = 4 3 3   

6 17. (a) ∵ a = 3, R = = 2 and n = 8 3

and

3( 2 −1) ∴ S(8) = 2 −1

∴

∴

=19

(c) ∵ a = 8, R =

∴ S(8) =

5 1 = and n = 6 10 2 6

  

11 16

19. (a) ∵ a = 28 and R =

∴

6 1 = − and n = 12 −18 3

S (∞) = ∴

=

and S(N) > 2000

117

28 1 1− 2 −18 2 =− 27 3

27  2 1 − −   3

81 5 

9



 

∴

3   5 S (∞) =    3 1 − −   5 =

265 720 =− 19 683

3 =1 1

14 1 = 28 2

− 25  3 3   (c) ∵ a = , R =  3  =−5 5 5 

  1 12  − 18 1 −  −     3    1 1 − −   3

18. (a) Let N be the number of terms of the given series.

S (∞) =

(b) ∵ a = 27, R =

=−13 120

∵ a = 1, R =

> 3650

= 56

8[1 −( −3) 8 ] 1 −( −3)

∴ S(12) =

(4 n −1)

4n – 1 > 16425 n > 7.002 ∴ 8 terms of geometric series must be taken.

−24 = –3 and n = 8 8

(d) ∵ a = –18, R =

3

4 −1

=765

 1 10 1 −     2  S(6) = 1 1− 2

S(N) > 3650 2

8

(b) ∵ a = 10, R =

1(3 N −1) > 2000 3 −1

20. (a)

3 8

 = 0.5555… 0.5 = 0.5 + 0.05 + 0.005 + 0.0005 + …

14

0.5 = 1 −0.1 5 = 9

S(10) =

2( 210 −1) 2 −1

= 2046 For the arithmetic sequence log 2, log 4, log 8, … a = log 2, d = log 2,

4  = 0.414 1414… (b) 0.41 = 0.4 + 0.014 + 0.000 14 + …

0.014 =0.4 + 1 −0.01 41 = 99 (c)

Arithmetic and Geometric Sequences and their Summation

S(10) =

10 (log 2 + 10 log 2) 2

= 55 log 2 ∴ The required sum = 2046

24. (a) (i) Common ratio =

(ii) S(n) =

a (10 n −1) 10 −1

=

Level 2 21. ∵ The equation ax2 +2bx + c = 0 has equal roots. ∴ ∆ =0 i.e. (2b)2 – 4ac = 0 b2 = ac

c b = b a

a (10 n −1) 9

(b) (i) T(2) – T(1) = log 10a – log a = log 10 = 1 T(3) – T(2) = log 100a – log 10a = log 10 = 1 … ∴ log a, log 10a, log 100a, … are in arithmetic sequence. (ii) S(n) =

∴ a, b, c are in geometric sequence.

10 a a

=10

 14  = 0.414 414… 0 .4 = 0.414 + 0.000 414 + 0.000 000 414 + …

0.414 = 1 −0.001 46 = 111

+55 log 2

n [2 log a + (n – 1)(1)] 2

= n log a +

n( n −1) 2

tan θcos θ = 2 sin θ

22.

tan θ cos θ = 2 sin2θ sin θ = 2 sin2θ sin θ (2 sinθ – 1) = 0 sin θ = 0 or

sinθ =

 θ = 30

or 150

25. (a) The weight loss in a month is a geometric sequence with a = 2 and R = 80% = 0.8. n− 1 kg ∴ In the nth month, weight loss = 2 ⋅0.8

1 2

23. (a) a ⋅ a2 ⋅ a3 ⋅ a3 ⋅ …⋅ a100 = a1 + 2 + 3 + … + 100

=a

100 (1+100 ) 2

=a 5050 (b) (2 + log 2) + (4 + log 4) + (8 + log 8) + … = (2 + 4 + 8 + …) + (log 2 + log 4 + log 8 + …) Then we have a geometric sequence 2, 4, 8, … and an arithmetic sequence log 2, log 4, log 8, … For the geometric sequence 2, 4, 8, …, a = 2, R = 2,

80

(b) Total weight loss =S (∞)



=

2 kg 1 −0.8

= 10 kg ∴ Her ultimate weight = (80 – 10) kg = 70 kg 26. ∵ The diameters of the semi-circles are in geometric sequence with a = 8 mm, R = 60% = 0.6 ∴ The lengths of the semi-circles are in geometric sequence with a =

8 π mm = 4π mm, R = 0.6 2

Maximum length =S (∞)

118

Certificate Mathematics in Action Full Solutions 5A

4π = mm 1 −0.6 =10 π mm

sin 2 α cos 2 α sin 2 α

27. (a) Common ratio =

= cos

(b) S(∞) =

2

S(∞) =

Side of the 3rd square = 8 cm 2

 1  = 16   cm  2 k- 1

 1     2 

∴ Side of the kth square =16  

geometric sequence with a = 16 cm, R =

3 1 = 9 3

9 1−

1 2

.

The perimeter of the squares are also in geometric sequence with a = 4 × 16 cm = 64 cm, R =

27 = 2

1

cm

(b) The lengths of each side of the squares are in

=1

=

2 cm

 1  = 16   cm  2

α

sin 2 α 1 − cos 2 α

28. (a) a = 9, R =

Side of the 2nd square = 8

3

5  1 9 1 −     3  S(5) =     121 = 1 9 1−

∴ S(10) =

1 2

.

  1 10  64 1 −      2   cm 1 1− 2

=62 ( 2 + 2 ) cm

3

∴ Error = S(5) – S (∞)

=

(c) The areas of the squares are in geometric sequence

121 27 − 2 9

with a = (16 × 16) cm2 = 256 cm2 and R =

1 =− 18

(b)

1 . 2

Total area of the infinite number of squares formed

 1 −  18  = × 100% Percentage error  27     2 

= =

a 1 −R 256 1−

1

cm 2

2

= 512 cm 2

= –0.412%

29. (a) Side of the 2nd square 2

=

2

 16   16    +   cm 2 2 

=8

2 cm

Side of the 3rd square 2

=

2

8 2  8 2       2  +  2  cm    

=8 cm Side of the 1st square = 16 cm

119

30. (a) (i) From 1st January 2010 to 31st December 2019, there are 20 half-years. The 1st deposit will amount to x(1.03)20. The 2nd deposit will amount to x(1.03)19. The 3rd deposit will amount to x(1.03)18. ∴ The required sum = $[x(1.03)20 + x(1.03)19 + x(1.03)18] 20 + 1.03 19 + 1.03 18 ) = $ x (1.03 (ii) The last deposit will amount to 1.03x. The total sum = $[x(1.03)20 + x(1.03)19 +… + 1.03x]

14

Arithmetic and Geometric Sequences and their Summation

1.03 x (1.03 20 −1) 1.03 −1 103 x (1.03 20 −1) =$ 3 =$

a= ∴

103 x (1.03 3

−1)

= 500 000

x =18 066

∩

(cor. to the

nearest integer) 31. ∵ a, b, c are in arithmetic sequence. ∴ b–a=c–b b2(c + a) – a2(b + c) = b2c + b2a – a2b – a2c = c(b2 – a2) + ab(b – a) = (b – a)[c(b + a) + ab] = (b – a)(cb + ca + ab) c2(a + b) – b2(c + a) = c2a + c2b – b2c – b2a = a(c2 – b2) + bc(c – b) = (c – b)[a(c + b) + bc] = (c – b)(ac + ab + bc) = (b – a)(cb + ca + ab) = b2(c + a) – a2(b + c) 2 2 ∴ a (b + c), b (c + a), c2(a + b) are in arithmetic sequence.

(c) Area of △OA2B1 =

πθ

A1 B1 = OA1 (b) (i)

=

Area of △OA3B2 =

∩

A2 B2 = OA2

1 (OB2)(OA3)sin θ 2 1 = (kcos θ )(kcos2θ )sin θ 2

Area of △OA4B3 =

θ

1 (OB3)(OA4)sin θ 2 1 = (kcos2θ )(OB3cos θ )sin θ 2 1 = (kcos2θ )(kcos2θ cos θ )sin 2 1 2 5 = 2 k cos θsin θ

(d) The areas of the triangles are in geometric sequence

πθ

180  πkθ cos θ = 180 

∩

1 (OB1)(OA2)sin θ 2 1 = (k)(kcos θ )sin θ 2

1 2 3 = 2 k cos θsin θ

180 

kπθ 180 

∩

1 2 = 2 k cos θ sin θ

32. (a) OA2 = OB1cos θ = kcos θ OA3 = OB2cos θ 2 = kcos θ ∩

∩

A1 B1 + A2 B2 + A3 B3 +...  kπθ       = 180  1 −cos θ kπ θ = 180  (1 −cos θ)

(b) The sum = 500 000 20

kπθ and R = cos θ . 180 

with a =

1 2 k cos θ sin θ and R = cos2θ . 2

∴ The sum to infinity of the series

πθ 180  πkθ cos 2 θ = 180 

A3 B3 = OA3

(ii) The sum is a geometric series with

80

120

Certificate Mathematics in Action Full Solutions 5A

1

=

2

k 2 cos θ sin θ

1 − cos 2 θ k 2 cos θ sin θ = 2 sin 2 θ 2 k cos θ = 2 sin θ k2 = 2 tan θ 8 8 8 (9) = (10 – 1) = (101 – 1) 9 9 9 8 8 8 88 = (99) = (100 – 1) = (102 – 1) 9 9 9 8 8 8 888 = (999) = (1000 – 1) = (103 – 1) 9 9 9

33. (a) 8 =

8 ∴ T(n) = 9 (10

n

−1)

(b) T(1) + T(2) + T(3) + … + T(n)

8 8 8 (101 – 1) + (102 – 1) + … + (10n – 1) 9 9 9 8 = [101 – 1 + 102 – 1 + … + 10n – 1] 9 8 = [101 + 102 + … + 10n – n] 9 =

 8 10 (10 n −1) −n   9 ( 10 − 1 )    n  8 10 (10 −1) =  −n  9 9    =

34. (a) (i) a4 = 1 + 2(1) + 2(2) + 2(3) = 13 a5 = 1 + 2(1) + 2(2) + 2(3) + 2(4) = 21 (ii) l = 2(n − 1) an = 1 + 2(1) + 2(2) + … + l = 1 + 2(1) + 2(2) + … + 2(n – 1) = 1 + 2[1 + 2 + … + (n – 1)]

 n( n −1)   2  

= 1 + 2

= 1 +n( n

121

− 1)

(b) (i) 1 = 2(1) – 1 5 = 2(1 + 2) – 1 11 = 2(1 + 2 + 3) – 1 19 = 2(1 + 2 + 3 + 4) – 1 ∴ L = 2(1 + 2 + 3 + … + n) – 1

 n( n +1)  –1 2  

= 2 =

n(n + 1) − 1

(ii) The terms in bn are in arithmetic sequence with a = an = 1 + n(n – 1) and d = 2. The number of terms = n ∴ The sum of terms in bn = an + (an + 2) + (an + 4) + … + L =

n [(1 + n(n – 1)) + (n(n + 1) – 1)] 2

3 =n

35. No Let the speed of the tortoise be v m/s, then that of Achilles be 5v m/s. ∵ The speed of Achilles is 5 times that of the tortoise. ∴ Distance travelled by Achilles is 5 times that of the tortoise. ∴ When Achilles reaches B, the tortoise move 2 m. i.e. BC = 2 m

1  m 5 

Similarly, CD = BC  =

2 m 5

∴ AB, BC, CD, … are in geometric sequence with a = 10 m and R =

1 . 5

∴ Total distance travelled by Achilles when he reaches the tortoise

10 =

1−

1

m

5

= 12.5 m ∴ The time needed =

12 .5 5 = s , which is finite. 5v 2v

36. (a) 1, 2, 3 or 2, 4, 6 or 3, 6, 9 (or any other reasonable answers) (b) By adding the corresponding terms of 1, 2, 3 and 2, 4, 6, we have 3, 6, 9, which is an arithmetic sequence with a common difference 3. The common difference of the new sequence is the sum of that of the former sequences. 37. (a) 1, 2, 4 or 1, 3, 9 or 1, 4, 16 (or any other reasonable answers)

14

Arithmetic and Geometric Sequences and their Summation

m1 + m2 a + d1 + a + 2d1 = n1 + n3 a + d 2 + a + 3d 2

(b) By multiplying the corresponding terms of 1, 2, 4 and 1, 3, 9, we have 1, 6, 36, which is a geometric sequence with a common ratio 6. The common ratio of the new sequence is the product of that of the former sequences.

Multiple Choice Questions (p. 216) 1.

Answer: B Let d be the common difference of the arithmetic sequence to be formed. The arithmetic sequence formed is: x, x + d, x + 2d, x + 3d, x + 4d, x + 5d, y ∵ The 7th term is also given by x + 6d. ∴ x + 6d = y d=

2.

3.

4.

7.

Answer: C Consider the geometric sequence: 1, 2, 4 and 1 + 10, 2 + 10, 4 + 10 = 11, 12, 14 which is not a geometric sequence. ∴ I is false. ∵ a, b , c are in geometric sequence.

b c = =R a b 10 b b = =R ∵ 10 a a 10 c c = =R and 10 b b ∴

Answer: C 10 ⋅ 102 ⋅ 103 ⋅ …⋅ 10n > 1050 101 + 2 + 3 + … + n > 1050 1 + 2 + 3 + … + n > 50

n( n +1) > 50 2

∴ 10a, 10b , 10c are in geometric sequence. ∴ II is true.

∴

5.

a +b a +b

Answer: C Consider the sequence which terms that are multiples of 7: 7, 14, 21, … Let n be the number of multiples of 7 less than 100. T(n) < 100 7 + (n – 1)(7) < 100 n < 14.29 ∴ The number of multiples of 7 less than 100 is 14. ∴ The number of positive integers less than 100 not divisible by 7 = 99 – 14 = 85

Answer: C T(7) = S(7) – S(6) = [3(7)2 – 2(7)] – [3(6)2 – 2(6)] = 133 – 96 = 37

n(n + 1) > 100 (n + 1)(n + 1) > 100 n + 1 > 10 or ∴ n > 9 or The smallest value of n is 10.

=

6.

x+y . 2

Answer: B d = n2 – n1 Common difference of the new arithmetic sequence = (3n2 + 1) – (3n1 + 1) = 3(n2 – n1) = 3d

a + a + 3d1 a + a + 4d 2

=1

y −x 6

∴ The 3rd arithmetic mean is

=

∵

n + 1 < –10 n < –11(rejected)

Answer: A Let d1 and d2 be the common differences of the arithmetic sequences respectively. Then b = a + 3d1 and b = a + 4d2

and

b a c b

=

b = R a

=

c = R b

∴ a, b, ∴ III is true. 8.

c are in geometric sequence.

Answer: C For I: 1  33   = 1 1  11 3   

and

 1 333  1 33 

   = 11 ≠ 1  111 11  

∴ I is not in geometric sequence.

80

122

Certificate Mathematics in Action Full Solutions 5A

2. For II:

( n −1)( n − 2) +1 ]th to 2 n( n −1) th 2

0.33 11 0.333 111 11 = = ≠ and 0.3 10 0.33 110 10

i.e. from [

∴ II is not in geometric sequence. For III: − 1   6   =−1 1  2 3   

,

1  12    =−1 − 1  2  6  

and

− 1   24    =−1 1  2 12   

If the denominator is 46, the terms are from [

( 46 −1)( 46 − 2) 46 ( 46 −1) th +1 ]th to 2 2

∴ III is in geometric sequence. 9.

i.e. from 991th to 1035th. For each denominator with value of n, the numerator runs from 1 to n – 1. By counting from the 991th term, the numerator of the 1001th term is 11.

Answer: D

1 10 1 xy = 100 xy =

2.

log x + log y = log xy

= –2 10. Answer: D 2

1 – sinQ30° = 1– cosR45° sinQ30° = cosR45°

n +1

1[( 9 ) −1] 2 9 −1 2 2 n +2 −1 1 + 9 + 9 + … +9 = (3 ) 2 4 2n 80 4 n +4 3 −1 = 80 =

11. Answer: B

12. Answer: D

S (∞) = 100

a = 100 1 −0.25 a = 75

HKMO (p. 217)

123

∵ sin 30° + sin230° + … + sinQ30° = 1– cosR45° ∴

sin 30  (1 − sin Q 30  ) = 1 − cos R 45   1 − sin 30 0.5(1 − sin Q 30  ) = 1 − cos R 45  0.5

1 = log 100

1.

the terms are from [1 + 2 + 3 +… + (n – 2) + 1]th to [1 + 2 + 3 +… + (n – 1)]th

Consider the sequence of the denominator. 2, 3, 3, 4, 4, 4, … , n, n, … ,n, … For each n, 1. there are (n – 1) terms.

Q

 1  1    =   2  2 2Q

 1   =    2

R=

2Q

 1     2   ∴

R

R