12[Anal Add Math CD]

1Additional Mathematics SPM Chapter 12

Penerbitan Pelangi Sdn. Bhd.

1. (a) d = 3(b) d = 2(d) d = 1(e) d = 0.2

2. (a) d = 5 2 = 3

x = 2 3 = 1 y = 8 + 3 = 11(b) d = 5 (1) = 4 x = 1 ( 4) = 3 y = 9 + ( 4) = 13(c) d = 1 12 = 12 x = 12

12 = 0

y = 32 + 12

= 2(d) d = 0.2 0.4 = 0.2 (x + 0.1) d = 0 x + 0.1 + 0.2 = 0 x = 0.3

3. (a) a = 5, d = 8 5 = 3 T51 = a + 50d = 5 + 50(3) = 155

(b) a = 2, d = 0 (2) = 2 T51 = a + 50d

= 2 + 50(2) = 98

(c) a = 1, d = 5 (1) = 4 T51 = a + 50d = 1 + 50(4) = 201

(d) a = 12 , d = 12

12 = 1 T51 = a + 50d = 12 + 50(1)

= 12 50

= 49 12

4. (a) a = 2, d = 6 2 = 4 Tn = 86 a + (n 1)d = 86 2 + (n 1)(4) = 86 (n 1)(4) = 84 n 1 = 21 n = 22(b) a = 1, d = 3 (1) = 2 Tn = 61 a + (n 1)d = 61 1 +(n 1)(2) = 61 n 1 = 602 = 30 n = 31

CHAPTER

12 Progressions

2 Additional Mathematics SPM Chapter 12


(c) a = x, d = (x + 2) x = 2 Tn = x + 50 a + (n 1)d = x + 50 x + (n 1)(2) = x + 50 n 1 = 502 = 25 n = 26

5. (a) a = 7, d = 10 7 = 3 S20 =

n2 [2a + (n 1)d]

= 202 [2(7) + (20 1)3] = 710

(b) a = 5, d = 7 (5) = 2

S20 = 202 [2(5) + (20 1)(2)]

= 10(10 38) = 480

(c) a = 1, d = 1.5 1 = 0.5

S20 = 202 [2(1) + (20 1)(0.5)]

= 10(11.5) = 115

(d) a = 14 , d = 34

14 = 12

S20 = 202 321

14 2 + (20 1)112 24

= 100

6. (a) T1 = S1 = 12 [2(1) + 1]

= 32(b) T2 = S2 S1 = 22 [2(2) + 1]

32 = 5 32 = 72

(c) T10 = S10 S9 = 102 [2(10) + 1]

92 [2(9) + 1] = 105 85.5 = 19.5(d) Tm 2 = Sm 2 Sm 3 = m 22 [2(m 2) + 1]

m 32 [2(m 3) + 1]

= m 22 (2m 3) m 32 (2m 5)

= (m 2)(2m 3)2 (m 3)(2m 5)2

= (m 2)(2m 3) (m 3)(2m 5)2 = (2m

2 7m + 6) (2m2 11m + 15)2 = 4m 92

7. (a) a = 2, d = 4 T6 = a + 5d = 2 + 5(4) = 22 T19 = a + 18d = 2 + 18(4) = 74 The sum between 5th term and 20th term = n2 (a + l)

= 142 (22 + 74) = 672(b) a = 5, d = 3 (5) = 2 T6 = a + 5d = 5 + 5(2) = 5 The sum between 5th term and 20th term = n2 [2a + (n 1)d]

= 142 [2(5) + (14 1)(2)] = 7(36) = 252(c) a = 8, d = 5 8 = 3 Sn =

n2 [2a + (n 1)d]

S19 = 192 [2 8 + (19 1)(3)]

= 361 S5 =

52 [2 8 + (5 1)(3)] = 10



The sum between 5th term and 20th term= 361 10= 371

8. (a) a = 3, d = 8 3 = 5

S20 = 202 [2(3) + (20 1)(5)]

= 10(101) = 1010

S4 = 42 [2(3) + (4 1)(5)]

= 2(21) = 42 The sum from 5th term to 20th term = 1010 42 = 968(b) a = 10, d = 6 10 = 4

S20 = 202 [2(10) + (20 1)( 4)]

= 560

S4 = 42 [2(10) + (4 1)( 4)]

= 16 The sum from 5th term to 20th term = 560 16 = 576(c) a = 9, d = 6 (9) = 3

S20 = 202 [2(9) + (20 1)(3)]

= 390

S4 = 42 [2(9) + (4 1)(3)]

= 18 The sum from 5th term to 20th term = 390 (18) = 408

9. (a) a = 15, d = 19 15 = 4

S14 = 142 [2(15) + (14 1)(4)]

= 574

S4 = 42 [2(15) + (4 1)(4)]

= 84 Sum of 10 terms = 574 84 = 490

(b) a = 21, d = 18 21 = 3

S13 = 132 [2(21) + (13 1)(3)]

= 132 (6)

= 39 S3 = 21 + 18 + 15 = 54 Sum of 10 terms = 39 54 = 15(c) a = 8, d = 6 (8) = 2

S14 = 142 [2(8) + (14 1)(2)]

= 70 S4 = (8) + (6) + (4) + (2) = 20 Sum of 10 terms = 70 (20) = 90

10. (a) d = 5 3 = 2 T11 = 3 a + 10d = 3 a + 10(2) = 3 a = 17

S10 = 102 [2(17) + (10 1)(2)]

= 80(b) d = 4 (1) = 3 T11 = 1 a + 10d = 1 a + 10(3) = 1 a = 29

S10 = 102 [2(29) + (10 1)(3)]

= 5(31) = 155

11. (a) a = 8, d = 4 (8) = 4

S10 = 102 [2(8) + (10 1)(4)]

= 5(20) = 100



(b) a = 18, d = 12 18 = 6

S10 = 102 [2(18) + (10 1)(6)]

= 90(c) a = 1.0, d = 1.4 1.0 = 0.4

S10 = 102 [2(1.0) + (10 1)(0.4)]

= 28

12. a = 2, d = 3 Sn = 155 n2 [2a + (n 1)d] = 155n2 [2(2) + (n 1)3] = 155

n2 (4 + 3n 3) = 155

n2 (1 + 3n) = 155 n + 3n2 = 310 3n2 + n 310 = 0 (3n + 31)(n 10) = 0 n = 313 , 10

Since n is the number of terms, then n = 10.

13. a = 5, Tn = 25 5 + (n 1)d = 25 (n 1)d = 30 ..............................1 Sn = 50n2 [2(5) + (n 1)d] = 50

n[10 + (n 1)d] = 100 10n + (n 1)dn = 100 .................2Substitute 1 into 2, 10n + 30n = 100 20n = 100 n = 5

14. T2 = 7a + d = 7 ...........................................1 T7 = 17a + 6d = 17 .......................................22 1, 5d = 10 d = 2

Substitute d = 2 into 1, a 2 = 7 a = 5

S20 = n2 [2a + (n 1)d]

= 202 [2(5) + (20 1)(2)] = 480

15. (a) T3 = 14 a + 2d = 14 .................................1 S10 = 185 102 [2a + (10 1)d] = 185 2a + 9d = 37 ............21 9, 9a + 18d = 126 .....................32 2, 4a + 18d = 74 .......................43 4, 5a = 52 a = 525 = 10 25

(b) Substitute a = 525 into 1,

525 + 2d = 14

2d = 14 525 = 185 d = 95(c) T20 = a + 19d = 525 + 191

95 2 = 525 +

1715 = 2235

16. a = 1, d = 2 Sn = 400 n2 [2a + (n 1)d] = 400

n2 [2(1) + (n 1)(2)] = 400 n(2 + 2n 2) = 800 2n2 = 800 n2 = 400 n = 20



17. a = p, d = 2 T6 = 140 a + 5d = 140p + 5(2) = 140 p = 140 + 10 = 150

18. a = 50, d = 2 Tn = 0 a + (n 1)d = 050 + (n 1)(2) = 0 n 1 = 25 n = 26Total distance covered = 26 400 = 10 400 m

19. (a) r = 3(b) r = 12

20. (a) a = 12 , r = 82

= 4 T6 = ar5

= 12 (4)5

= 512

(b) a = 8, r = 48 = 12 T6 = ar5

= (8)1 12 25

= 14

(c) a = 0.6, r = 1.20.6 = 2 T6 = ar5 = 0.6 (2)5 = 19.2

21. (a) a = 10, r = 510 = 12

Tn = 5128

arn 1 = 5128 101 12 2

n 1 = 5128

1 12 2n 1

= 1256 = 1 12 2

8

n 1 = 8 n = 9

(b) a = 2, r = 62 = 3 Tn = 4374 arn 1 = 4374 (2)(3)n 1 = 4374 3n 1 = 2187 = 37 n 1 = 7 n = 8

22. (a) a = 7, r = 147 = 2

Sn = a1 rn 1r 1 2

S10 = 71 210 12 1 2

= 7161

(b) a = 4, r = 2 4 = 12

S10 = a11 rn1 r 2

= ( 4)3 1 112 2

10

1 12

4 = 831 1 12 2

104

= 7 127128



23. a = 8, r = 248 = 3Sum between 5th term and 10th term= S9 S5= 81 3

9 13 1 2 8135 13 1 2

= 4(39 1) 4(35 1)= 78 728 968= 77 760

24. a = 1, r = 12

S9 S3 = 13 1 1 12 2

9

1 1 12 2

4 13 1 1 12 2

3

1 1 12 2

4 = 23 31 1

12 294 23 31 1

12 234

= 21256

25. r = 105 = 2 T8 = 5 ar7 = 5 a(2)7 = 5 a = 527

= 5128Sum of 7 terms before 5= S7= 5128 1

27 12 1 2= 635128

26. a = 5, r = 105 = 2 Sn = 155

a1 rn 1r 1 2 = 155

51 2n 12 1 2 = 155

2n 1 = 31 2n = 32 = 25 n = 5

27. a = 10, r = 510 = 12 Sn =

1558 a1 1 r

n1 r 2 =

1558

103 1 112 2

n

1 12

4 = 1558 2031 1 12 2

n4 = 1558

1 1 12 2n = 155160

1 12 2n = 1 155160

= 5160 = 132

= 1 12 25

n = 5

28. (a) a = 18, r = 918 = 12 S

= a1 r

= 181 12

= 36

(b) a = 12, r = 412 = 13 S

= a1 r

= 121 13

= 121 32 2 = 18



29. a = 10, S = 20

a1 r = 20

101 r = 20

1 r = 12 r = 1 12 = 12

30. r = 13 , S = 40

a1 r = 40

a1 13

= 40

a = 40 23 = 803

31. (a) p1 = 2 pp

p2 = 2 p p2 + p 2 = 0 (p + 2)(p 1) = 0 p = 2, 1 Since p = 1 does not form the geometric

progression, then p = 2.

(b) r = p1 = 21 = 2

32. 1, 2, 4, a = 1, r = 2

S20 = a1 rn 1r 1 2

= 11 220 12 1 2

= 1 048 575

33. 2, 6, 18, a = 2, r = 3S10 = 21 3

10 13 1 2 = RM59 048

34. T2 = 10 ar = 10 ................................................1 T5 = 80 ar 4 = 80 ..............................................221

, ar 4ar = 8010

r3 = 8 r = 2Substitute r = 2 into 1,a = 5T3 = ar2 = 5 (2)2 = 20

35. T2 + T3 = 15 ar + ar2 = 15 a(r + r2) = 15 ....................................1a = 20 ................................................2Substitute 2 into 1, 20(r + r2) = 15 r + r2 = 34 4r2 + 4r 3 = 0(2r 1)(2r + 3) = 0 r = 12 ,

32Since r > 0, then r = 12 .

36. (a) S = 16

a1 r = 16 a = 16(1 r) ......................1 T2 = 4 ar = 4 ..........................................2 Substitute 1 into 2, 16(1 r)(r) = 4 16r 16r 2 = 4 16r2 16r + 4 = 0 4r2 4r + 1 = 0 (2r 1)2 = 0 r = 12 Substitute r = 12 into 1,

a = 1611 12 2 = 8 Therefore, the common ratio = 12 and the first

term = 8.



(b) T5 = ar4

= 81 12 24

= 12 37. (a) 2, 4, 8, 16,

Area of the 2nd circleArea of 1st circle =

4222

= 4

Area of the 3rd circleArea of the 2nd circle =

8242

= 4

Since they have same common ratio, r = 4, the area of the circle forms a geometric progression.

(b) Area of the 7th circle = R2 where R is the radius of the 7th circle. R = T7 = ar6

= 2 1 42 26

= 27 m Area of the 7th circle = R2 = (214) = 16 384 m2

1. d = (5) (8) = 3 S4 = 96n2 [2a + (n 1)d] = 9642 [2a + (4 1)3] = 96 2a + 9 = 48

2a = 39 a = 392Therefore, the four consecutive terms are392 ,

452 , 512 ,

572 .

2. p 2 = q p 2p q = 2

3. (a) (2k 9) (k 2) = (k + 2) (2k 9) 2k 9 k + 2 = k + 2 2k + 9 k 7 = k + 11 2k = 18 k = 9(b) a = k 2 = 9 2 = 7

d = (2k 9) (k 2) = (18 9) (9 2) = 9 7 = 2 S10 =

n2 [2a + (n 1)d]

= 102 [2(7) + (10 1)(2)] = 5(32) = 160

4. (a) a = 4 d = 6 ( 4) = 2(b) S34 S4 = 342 [2( 4) + (34 1)(2)]

42 [2( 4) + (4 1)(2)]

= 17(74) 2(14) = 1230

5. T3 = 1 + 3m a + 2d = 1 + 3ma + 2(3) = 1 + 3m a = 3m 5 ...............................1 S5 = 5m + 452 (2a + 4d) = 5m + 4

5a + 10d = 5m + 4 5a + 10(3) = 5m + 4 5a = 5m 26 ........................2Substitute 1 into 2, 5(3m 5) = 5m 26 15m 25 = 5m 26 15m 5m = 26 + 25 10m = 1 m = 110

6. Let Sn . 0 n2 [2a + (n 1)d] . 0, where a = 51 and d = 3 n2 [2(51) + (n 1)(3)] . 0 n2 (102 + 3n 3) . 0 n2 (3n 105) . 0

32 n(n 35) . 0



0 35

n 0, n . 35Therefore, the minimum number of terms is 36.

7. p3x + 2 = 3x + 210

p = (3x + 2)2

10

8. 0.181818 = 0.18 + 0.0018 + 0.000018 + a = 0.18, r = 0.00180.18 = 0.010.181818 = a1 r = 0.181 0.01 = 211

9. (a) a = 3, r = 123 = 4 Common ratio = 4(b) Sn = 65 535 a1 r

n 1r 1 2 = 65 535 31 4

n 14 1 2 = 65 535 4n = 65 536 = 48 n = 8

10. (a) T2 = 3 ar = 3..........................................1 T4 =

34 ar3 = 34 .......................................2

21

, ar 3ar = 343

r2 = 14 r = 12 Since r . 0, then r = 12 .

(b) Substitute r = 12 into 1,

a = 312 = 6 S

= a1 r

= 61 12

= 12

11. (a) T2 = 1 ar = 1 ...........................................1 T3 + T4 = 2 ar2 + ar3 = 2 ar(r + r2) = 2 ...............................2 Substitute 1 into 2, 1(r + r2) = 2 r + r2 = 2 r2 + r 2 = 0 (r + 2)(r 1) = 0 r = 2, 1 Since r 0, then r = 2. Substitute r = 2 into 1, a(2) = 1 a = 12

(b) S10 = a1 1 rn

1 r 2 = 12 3

1 (2)101 (2) 4 = 16 [1 (2)

10]

= 10236 = 170.5

12. (a) 1, 3, 5 a = 1, d = 2 Tn = 401 a + (n 1)d = 401 1 + (n 1)(2) = 401 (n 1) = 4002 n = 201 The height is 201 3 = 603 cm

10

Additional Mathematics SPM Chapter 12


(b) Sn = n2 [2a + (n 1)d]

S201 = 2012 [2 1 + (200)2]

= 2012 (402) = 201 201 = 40 401

13. (a) 1st 2nd 3rd

x 23 x49 x

y 23 y49 y

A1 = Area of the 1st rectangle = xy

A2 = Area of the 2nd rectangle = 49 xy

A3 = Area of the 3rd rectangle = 1681 xy

A2A1 =

49 xyxy = 49

A3A2 =

1681 xy49 xy

= 1681 94

= 49 Since the ratios are the same, then the areas of

the rectangle form a geometric progression with common ratio = 49 .

(b) (i) A1 = xy = 4 9 = 36 A2 =

49 xy

= 49 36 = 16

A3 = 1681 xy

= 1681 36

= 7 19

Tn = arn 1

3 1381 = 36149 2

n 1

25681 = 36149 2

n 1

64729 = 149 2

n 1

1 49 23 = 1 49 2

n 1

n 1 = 3 n = 4 Therefore, the fourth rectangle has area

of 3 1381 m2.

(ii) S = a1 r

= 361 49

= 36 95 = 3245 m

2

14. (a) x, x + y, x + 2y, a = x, d = y given T6 = 200 a + 5d = 200 x + 5(y) = 200 x + 5y = 200 .............................1 and S12 = 2520 n2 [2a + (n 1)d] = 2520

122 (2x + 11y) = 2520 2x + 11y = 420..............2 1 2, 2x + 10y = 400 ..............3 2 3, y = 20 Substitute y = 20 into 1, x + 5(20) = 200 x = 100

(b) Tn of Encik Ali = Tn of Encik Tan 100 + (n 1)20 = 200 + (n 1)10 100 + 20n 20 = 200 + 10n 10 10n = 110 n = 11

11



15. (a) 10x20x = 12

5x10x = 12

Therefore, it is a geometric progression.(b) The sequence is a geometric progression since it

has a common ratio of 12 .

(c) S = a1 r

40 = 20x1 12

40 = 40x x = 1

16. (a) 10 (h + k) = (2h 1) 10 10 h k = 2h 11 3h = 21 k h = 7 k3

(b) Given k = 3, h = 7 33 = 6 h + k = 6 + 3 = 9 Common difference = 10 9 = 1

1. (a) a = 24, d = 4 T10 = a + 9d = 24 + 9( 4) = 12

(b) S30 = 302 [2(24) + (30 1)(4)]

= 15(68) = 1020

2. 7, 9, 11, , 199a = 7, d = 2 Tn = 199a + (n 1)d = 1997 + (n 1)2 = 199 n 1 = 1922 = 96 n = 97

S97 = 972 (7 + 199)

= 9991

3. (3 x) (20) = 16 (3 x) 3 x + 20 = 16 + 3 + x 17 x = 13 + x 2x = 30 x = 15

4. x 2x2 = (3x 10) x x 2x2 = 2x 10 2x2 + x 10 = 0(2x + 5)(x 2) = 0 x = 52 , x = 2Since x . 0, then x = 2.

5. Tn = 15 3na = T1 = 15 3(1) = 12T10 = 15 3(10) = 15

Sn = n2 (a + l)

S10 = 102 [12 + (15)]

= 5(3) = 15

6. 2y + 3 2x = 5y + 1 (2y + 3) = 5y + 1 2y 3 2y + 3 2x = 3y 2 3y 2y = 3 2x + 2 y = 5 2x

7. (a) T10 = ar9

= 1 12 2(2)9 = 256

(b) S10 = a1 1 rn

1 r 2

= 1 12 231 (2)101 (2) 4

= 170 12

12



8. r = 13 S

= 30

a1 r = 30a

1 13 = 30

a23 = 30

a = 30 23 = 20

9. x2 = 12x

x2 = 1 x = 1Since x . 0, then x = 1.

10. Tn = 729 arn 1 = 729

1 13 2(3)n 1 = 729 (3)n 1 = 2187 = (3)7 n 1 = 7 n = 8

11. 5x 3x = 7x + 275x 3

(5x 3)2 = 7x2 + 27x 25x2 30x + 9 = 7x2 + 27x 18x2 57x + 9 = 0 (18x 3)(x 3) = 0 x = 318 , 3

Since x is an integer, then x = 3.

12. a = 4, r = 24 = 12S

= a1 r

= 41 12

= 8

13. 1t = 0.0833333 = 0.08 + 0.003 + 0.0003 + 0.00003 + = 0.08 + a1 r = 0.08 + 0.0031 0.1 = 0.08 + 0.0030.9 = 112 t = 12

14. 0.090909...= 0.09 + 0.0009 + 0.000009 + = a1 r= 0.091 0.01= 111

15. 1p = 0.166666 = 0.1 + 0.06 + 0.006 + 0.0006 + = 0.1 + a1 r = 0.1 + 0.061 0.1 = 0.1 + 0.060.9 = 16Therefore, a = 0.06, b = 0.006 and p = 6.

16. (a) p = q + 4 + 45 + 425 +

r = 454

= 15 4q =

15 q = 20

(b) a = 20, r = 15 p = S

= a1 r = 20

1 15 = 20 54 = 25

13



17. T1 + T2 = 28 a + ar = 28 a(1 + r) = 28 .......................................1 T2 + T3 = 14 ar + ar2 = 14a(r + r2) = 14 ....................................221

, a(r + r2)a(1 + r) = 1428

r(1 + r)1 + r = 12

r = 12

Substitute r = 12 into 1, a11 12 2 = 28 a = 28 2 = 56S6 = a11 r

61 r 2

= 563 1 1 12 2

6

1 1 12 2

4 = 563 1 1

12 26

32

4 = 36.75

18. Sn = n(6 n)(a) T3 = S3 S2 = 3(6 3) 2(6 2) = 9 8 = 1(b) S10 S5 = 10(6 10) 5(6 5) = 40 5 = 45

19. T10 = 7T1a + 9d = 7a 6a = 9d a = 32 d

20. a, a + d, a + 2d, a + 3d, a + 4d, S20 = 570202 [2a + (20 1)d] = 570 2a + 19d = 57 ................................1

Sum of even numbers: S10 = 300102 [2(a + d) + (10 1)(2d)] = 300 2a + 2d + 18d = 60 2a + 20d = 60 ..................22 1, d = 3Substitute d = 3 into 1,2a + 19(3) = 57 2a = 57 57 a = 0T10 = a + 9d = 0 + 9(3) = 27

21. x2x + 3 = x 2x

x2 = (x 2)(2x + 3) = 2x2 x 6 x2 x 6 = 0(x 3)(x + 2) = 0 x = 3, 2

When x = 3, the sequence is 9, 3, 1 and r = 39 = 13When x = 2, the sequence is 1, 2, 4 and

r = 21 = 2

Since 1 r 1 for S to exist,

then r = 13 .

S = a1 r

= 91 13

= 923 = 9 32 = 272

22. (a) 200, 180, 160, Tn = 0 a + (n 1)d = 0 200 + (n 1)(20) = 0 n 1 = 20020 = 10 n = 11

14



(b) Total distance travelled = 112 [2(200) + 10(20)]

= 112 (400 200) = 1100 m

23. (a) Progression I: 1, 1 + d, 1 + 2d, Progression II: A, A + 2, A + 4, A + 6 + 1 = A + 6 A = 5 Therefore, the first term of the arithmetic

progression II is 5.(b) S4 = 22 42 [2(1) + (4 1)d] = 22 2 + 3d = 11 3d = 9 d = 3

24. (a) (i) T6 = 15 a + 5d = 15 .........................1 T10 = 23 a + 9d = 23 ........................2 2 1, 4d = 8 d = 2 Substitute d = 2 into 1, a + 10 = 15 a = 5 (ii) S15 =

152 [2(5) + (15 1)(2)]

= 152 (38) = 285(b) Tn = 405 a + (n 1)d = 405 5 + (n 1)(2) = 405 (n 1)(2) = 400 n 1 = 200 n = 201 Since n is an integer, this shows that 405 is a term

in the progression.

25. (a) 98, 96, 94, a = 98 cm, d = 2 cm T20 = a + 19d = 98 + 19(2) = 60 cm

(b) 2S19 + T20 = 23 192 [2(98) + 18(2)4 + 60 = 19(160) + 60 = 3100 cm

26. (a) Particle P: 10, 12, 14, 16, Distance travelled by P = distance travelled by Q + 100 n2 [2 10 + (n 1)(2)] = 9n + 100

n2 (20 + 2n 2) = 9n + 100

n2 (18 + 2n) = 9n + 100 9n + n2 = 9n + 100 n2 = 100 n = 10 m s1 Therefore, the velocity of P is 10 m s1.(b) Distance travelled by P = 9 10 + 100 = 190 m

27. (a) 2, 1.8, 1.6, 1.4, a = 2, d = 0.2 Tn = 0 a + (n 1)d = 0 2 + (n 1)( 0.2) = 0 n 1 = 2 0.2 = 10 n = 11 Time taken is 11 seconds.

(b) S11 = 112 [2(2) + 10(0.2)]

= 112 (4 2) = 11 cm

28. (a) 2, 4, 6, 8, (height) a = 2, d = 2 T25 = a + 24d = 2 + 24(2) = 2 + 48 = 50 Therefore, the height of 25th edge is 50 cm.

(b) A1 = 12 2 (2 + 4)

= 6

A2 = 12 2 (4 + 6)

= 10

15



A3 = 12 2 (6 + 8)

= 14 a = 6, d = 4 The sum of the areas for the first 10 trapeziums

= 102 [2(6) + 9(4)] = 240 cm2

29. (a) (i) T2 = x2 ar = x2 ................................1 T3 =

x 416 ar2 = x

416 .............................2

21

, ar2ar = x416x2

r = x 216

(ii) r = x 216 = 1 or r =

x 216 = 0 x2 = 16 x = 0 x = 4 Therefore, the values of x are 0, 4, 4.(b) (i) 5x (x + 5) = (8x 3) 5x 4x 5 = 3x 3 x = 2 (ii) When x = 2, x + 5 = 7 5x = 10 8x 3 = 13 \ d = 3 a = T4 = 13 + 3 = 16 S10 =

102 [2(16) + 9(3)] = 5(32 + 27) = 5(59) = 295

30. (a) a = T1 = 2 n = 10 S10 = 155 102 [2(2) + 9d] = 155 4 + 9d = 31 9d = 27 d = 3

The longest part = T10 = a + 9d = 2 + 9(3) = 29(b) S10 S5 =

102 [2(2) + 9(3)] 52 [2(2) + 4(3)]

= 5(31) 52 (16) = 115

31. (a) 2, 6, 18, 54, The base is a geometric progression with a = 2, r = 3 T10 = ar9 = 2(3)9 = 39 366 The value of the 10th term = 39 3662 = 1 549 681 956(b) (i) 4x x = 3x 7x 4x = 3x Therefore, the indices form an arithmetic

progression. (ii) a = x, d = 3x S10 =

102 [2x + 9(3x)] = 5(2x + 27x) = 145x

32. a = 5, r = 105 = 2 Sn . 1000a1 rn 1r 1 2 . 1000512n 12 1 2 . 1000 2n 1 . 200 2n . 201 log10 2n . log10 201 n log10 2 . log10 201

n . log10 201log10 2

n . 7.65Therefore, the least number of n = 8.

33. (a) G.P.: T1 = a = 3, r A.P.: T1 = a = 3, d = r T5 of the geometric progression = 48 ar4 = 48 3r4 = 48 r4 = 16 r = 2 Since r . 0, then r = 2.

16



(b) Sn of A.P. = T4 of G.P. n2 [2(3) + (n 1)2] = 3 (2)

3

n2 (6 + 2n 2) = 24

n2 (4 + 2n) = 24 2n + n2 = 24 n2 + 2n 24 = 0 (n + 6)(n 4) = 0 n = 4, 6 Since n is the number of terms, therefore n = 4.

1. a = 121d = 116 121 = 5Let Tn 0 a + (n 1)d 0 121 + (n 1)(5) 0 (n 1)(5) 121 n 1 . 1215 n . 1215 + 1 n . 25.2Therefore, the first negative term is 26th term.

2. 8, 10, 12, , 200a = 8, d = 10 8 = 2Let Tn = 200 a + (n 1)d = 200 8 + (n 1)(2) = 200 (n 1)(2) = 200 8 n 1 = 1922 = 96 n = 97Sum of all even numbers, Sn =

n2 [2a + (n 1)d]

S97 = 972 [2(8) + 96 2]

= 10 088

3. 3, 9, 27, a = 3, r = 3Let Sn . 500

a1 r n 1r 1 2 . 500

31 3n 13 1 2 . 500

3n 1 . 500 23 3n . 10003 + 1

3n . 10033 log10 3n . log10

10033 n log10 3 . log10 1 10033 2

n . log10 1003 log10 3log10 3

n . 5.29Therefore, 6 days are required to produce more than 500 cells.

4. 200, 400, 800, It is a G.P. with a = 200 and r = 2.Let Sn . 1 000 000 a1 r

n 1r 1 2 . 1 000 000 2001 2

n 12 1 2 . 1 000 000 2n 1 . 5000 2n . 5001 n log10 2 . log10 5001

n . log10 5001 log10 2 n . 12.29Hence, in 13 days, all the money will be used up.

5. T4 = 10 a + 3d = 10 ............................1 T10 = 34 a + 9d = 34 ............................22 1, 6d = 24 d = 4Substitute d = 4 into 1, a + (3 4) = 10 a = 2Let Tn = 306 a + (n 1)d = 306 2 + (n 1)(4) = 306 (n 1)(4) = 308 n 1 = 77 n = 78Since n is an integer, therefore 306 is a term in the progression.

17



6. Circumferences:2pr, 2p(r + 1), 2p(r + 2), T2 T1 = 2p(r + 1) 2pr = 2pT3 T2 = 2p(r + 2) 2p(r + 1) = 2pr + 4p 2pr 2p = 2pSince T2 T1 = T3 T2, therefore the circumference of the circles are A.P..

Areas:pr2, p(r + 1)2, p(r + 2)2, T2 T1 = p(r + 1)2 pr2 = p(r2 + 2r + 1) pr2 = pr2 + 2pr + p pr2 = 2pr + pT3 T2 = p(r + 2)2 p(r + 1)2 = p(r2 + 4r + 4) p(r2 + 2r + 1) = pr2 + 4pr + 4p pr2 2pr p = 2pr + 3pSince T2 T1 T3 T2, therefore the area of the circles are not A.P..For circumferences,a = 2pr, d = 2pThe total circumferences of the first 5 circles

= 52 [2(2pr) + 4(2p)]

= 52 (4pr + 8p)

= 52 4p(r + 2)= 10p(r + 2) units

7. (a) 10, 13, 16, 19, are the radii of the sectors in A.P..

a = 10, d = 3 Let the radius of the nth sector = Tn. Tn = 10 + (n 1)3 = 7 + 3n Therefore, the radius of the nth sector is 7 + 3n.(b) Given the area of the nth sector = 512p cm2 12 r

2q = 512p

12 (7 + 3n)21 p4 2 = 512p

(7 + 3n)2 = 512p 8p

= 4096 7 + 3n = 64 3n = 57 n = 19

(c) 101 p4 2, 131p4 2, 161

p4 2, are the arc lengths of the sectors in A.P. where a = 10p4 =

5p2 and d = 3p4 .

The sum of the first 20 arc lengths

= n2 [2a + (n 1)d]

= 202 32152 p2 + 191

3p4 24 = 1015p + 574 p2 = 101 774 p2 = 3852 p cm

8. (a) r, r + 1, r + 2, are the radii of the semicircles which form an A.P..

a = r cm, d = 1 cm p(r), p(r + 1), p(r + 2), are the circumferences

of the semicircles which form an A.P.. a = pr, d = p Given the total circumferences of the first 8

semicircles = 84p n2 [2a + (n 1)d] = 84p

82 (2pr + 7p) = 84p 2pr + 7p = 21p 2r + 7 = 21 2r = 14 r = 7 Therefore, the radius of the smallest semicircle is

7 cm.

(b) Let Sn = 45p n2 [2a + (n 1)d] = 45p n2 [2pr + (n 1)p] = 45p n2 [2p(7) + (n 1)p] = 45p n2 (14 + n 1) = 45 n2 (13 + n) = 45 13n + n2 = 90 n2 + 13n 90 = 0 (n + 18)(n 5) = 0 n = 5, n = 18 is ignored since n . 0. Hence, the number of semicircles is 5.

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9. (a) 10, 12, 14, 16, are the distance, in m, travelled per second of the particle P which form an A.P..

a = 10, d = 2 The distance travelled by particle P after t

seconds, St =

t2 [2(10) + (t 1)2]

= t2 (20 + 2t 2)

= t2 (18 + 2t) = t(9 + t) m The distance travelled by particle Q in t seconds

8t m. Thus, t(9 + t) + 8t = 60 9t + t2 + 8t 60 = 0 t2 + 17t 60 = 0 (t 3)(t + 20) = 0 t = 3, t = 20 is ignored since t > 0.(b) Distance travelled by particle P = 3(9 + 3) = 36 m

10. (a) Given T30 = 99 a + 29d = 99.............................1 S3 = 45 32 (2a + 2d) = 45 3a + 3d = 45 a + d = 15 .........................2 1 2, 28d = 84 d = 8428 = 3 Substitute d = 3 into 2, a + 3 = 15 a = 12 Therefore, the length of the shortest part is

12 cm.

(b) p = S30 = 302 [2(12) + 29(3)] = 15(24 + 87) = 1665

11. Given T2 = 5 ar = 5 ....................1

T5 = 58

ar4 = 58 ..................2

21

, ar4ar =

585

r3 = 18 r = 12Substitute r = 12 into 1, a2 = 5 a = 10

S = a1 r

= 101 12

= 20

12. Given S = 120

a1 r = 120

801 r = 120

1 r = 80120 = 23 r = 1 23 = 13Let Sn . 118 a1 1 r

n 1 r 2 . 118

803 1 113 2

n

1 13

4 . 118 1 1 13 2

n . 11880

23 1 1 13 2

n . 5960

1 5960 . 113 2

n

160 . 113 2

n

log10 1 160 2 . log10 113 2

n

log10 1 log10 60 . n (log10 1 log10 3) 0 log10 60 . n (log10 3) . n log10 3

19



log10 60log10 3 n

3.727 nHence, the minimum number of terms is 4.

13. (a) 5, 2.5, 1.25, are the heights bounced by the yoyo in the subsequent processes which is a G.P. with a = 5, r = 12 .

At the fifth bounce = T5 = ar4

= 51 12 24

= 516 The distance from the boys finger

= 10 516 = 9 1116 cm

(b) S = a1 r

= 51 12

= 10 S

= 10 is the total distance bouncing up only.

Since the distance bouncing up and coming down are the same, then the total distance travelled

= 2 10 = 20 cm

14. For the heights: 20, 18, 16, which is an A.P. with T1 = 20 and

d = 2Height of sixth cylinder, T6 = a + 5d = 20 + 5(2) = 20 10 = 10For the diameters:10, 5, 52 , which is a G.P. with a = 10, r =

12Diameter of sixth cylinder, T6 = ar5

= 101 12 25

= 1032 = 516The area of the circle for sixth cylinder= p1 d2 2

2

= p1 516 2 22

= 251024p cm2

The volume of the sixth cylinder= 1 251024p2(10)

= 125512 p cm3

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