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4. nth term = a + (n − 1)d 11. T = ar = 3
2
Second term = 8k + d = k − 8 a = 3
−8 − 7k = d ……(1) r
Third term = 8k + 2d = 4 − k ……(2) 1 2
2
3
Substitute equation (1) into equation (2): r = 27
8k + 2(−8 − 7k) = 4 − k 1 – r 2
2
k − 6k − 20 = 0 27r – 27r + 6 = 0
2
2
k = −2.39 and k = 8.39
2 1
[3] r = ,
5. (a) C (3) + C (3) (x) + C (3) (x) + C (3) (x) 3 3 3
5
2
5
5
3
5
5
2
4
9
0 1 2 3 a = , 9
243 + 405x + 270x + 90x 2 [3]
3
2
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[2]
3 1 2
2 1 2
1 1 2
4
(b) C (4) + C (4) 4 –x + C (4) 3 –x 2 + C (4) 2 –x 3 12. (a) 8 (2a + (8 − 1)d) = 2 3 5 (2a + (5 − 1)d)
5
5
5
5
5
0 2 2 2 2 2
1024 − 640x + 160x − 20x
2
3
[2] a = 4d
n 3 n [2]
2
6. (a) 1 n + 8 = (2a + (n – 1)d) (b) T = a + (20 − 1)d
2 4 2 20 = 4d + 19d
3 n + 8 = 2a + nd – d = 23d
4
3 n = nd [1]
4 13. S = a(1 – r )
3
3
d = 3 1 – r
4 = 1.925
2a – d = 8
3
2a – = 8 a = 1.925(1 – r)
3
4 (1 – r )
a = 35 a
8 [3] S = 1 – r
∞
(b) S − S = 2.2
20 7
35
20
3
24
= 3 1 1 2 + (20 – 1) – a = 2.2(1 – r)
2
8
4
2
= 2.2(1 – r)
7
35
3
24
2
= 3 1 1 2 + (7 – 1) 1.925(1 – r) r = 0.5
(1 – r )
3
4
2
8
= 1469
8 [3] a = 2.2(1 − 0.5)
= 1.1
7. (a) C (2x) + C (2x) (−4) + C (2x) (−4) [4]
2
3
5
5
5
5
4
0 1 2
32x − 320x + 1280x 10 2 5 1 5
3
5
4
1
[2] 14. C (ax ) – 3x 22 = −8064
5
(b) (−320x )(−1) + (1280x )(3x) = 4160x 1
4
3
4
1
5
[2] 252a – 243 2 = −8064
–1
8. C (3x) + 1 2 3 = −7654.5 a = 6
9
6
3 2x 2 [4] [2]
9. 2 = 64 15. (a) Perimeter = 176 cm, 132 cm, 99 cm
n
n − 1
n = 6 nth term = ar
6 C (2) (ax) = −576x Second term = ar = 176r = 132
5
1 132
192ax = −576x r =
a = −3 116
4
2
6 C (2) (−3x) = 2160x = 3 4
2
2 b = 2160 3 4
[6] Fifth term (perimeter) = 176 1 2
4
10. S = 1 891
∞ –1 = cm
1 – 1 2 16 [3]
3
= 3
4 [2]
Answers 195
Answers Add Math.indd 195 14/03/2022 12:29 PM