Page 43 - ACE YR IGCSE A TOP APPR' TO MATHS

4 (a) T = n 2 [2] = 8 + (n – 1)(6)
n
(b) T = n + 1 [2] = 6n + 2
2
n
(c) T = 2(n + 1) Total perimeter
2
n
= 2n + 2 [2] = [6(1) + 2] + [6(2) + 2] + [6(3) + 2]… +
2
5 (a) 2a = 2 [6(49) + 2] + [6(50) + 2]
3a + b = 6 = 6(1) + 6(2) + 6(3)… + 6(49) + 6(50) +
a + b + c = 3 (2 × 50)
a = 1, b = 3, c = –1 = 6(1 + 2 + 3 + … + 49 + 50) + 100
T = n + 3n – 1 [3] = 6(1275) + 100 = 7750 [4]
2
n
(b) n + 3n – 1 = 269 9 (a)
2
n + 3n – 270 = 0 Diagram 5 6
2
(n – 15)(n + 18) = 0 Number of dots 21 28
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n = 15 or n = –18 (invalid) [2] Number of lines 45 63
6 (a) 2a = 4 Number of triangles 25 36
3a + b = 3
a + b + c = –1 [3]
a = 2, b = –3, c = 0 (b) (i) 2a = 1
T = 2n – 3n 3a + b = 3
2
n a + b + c = 3
2, 5 ,8, 11 1
T = a + (n – 1)d a = 2
n
= 2 + (n – 1)(3) 3
= 3n – 1 b = 2
2
T = 2n – 3n [3] c = 1
n
3n – 1
2
2
(b) T = 2(10) – 3(10) T = 1 2 n + 3 n + 1 [3]
2
n
10
3(10) – 1
2
= 170 [1] (ii) 1 n + 3 n + 1 = 136
2
2
29 n + 3n – 270 = 0
2
2
(c) 2n – 3n > 10 (n – 15)(n + 18) = 0
3n – 1
n = 15 or n = –18 (invalid)
2n – 3n > 30n – 10 (c) (i) 2a = 3 [2]
2
2n – 33n + 10 > 0 3a + b = 6
2
n < 0.309 (invalid) or n > 16.19 a + b + c = 3
n = 17 [2]
7 (a) T = 4 – 16 = 240 [1] a = 3
4
5
(b) 4, 7, 10, 13 2
3
a + (n – 1)d b =
= 4 + (n – 1)(3) c = 0 2
= 3n + 1
T = 4 n – 1 – (3n + 1) T = 3 n + 3 n [3]
2
n
T = 4 n – 1 – 3n – 1 [2] n 2 2
n
(c) T = 4 10 – 1 – 3(10) – 1 (ii) T = 3 (10) + 3 (10)
2
10
= 4 – 31 10 2 2
9
= 262113 [2] = 165 [1]
8 (a) a = 70 (d) T = n
2
n
b = 96 [1] T = 20 2
20
(b) 2a = 4 = 400 [2]
3a + b = 10 (e) Sequence of perimeter: 3, 6, 9, 12
a + b + c = 6 T = a + (n – 1)d
n
a = 2, b = 4, c = 0 T = 3 + (n – 1)(3)
n
t = 2n + 4n [3] T = 3n
2
n
(c) 390 = 2n + 4n 3n = 459
2
n + 2n – 195 = 0 n = 153 [2]
2
(n – 13)(n + 15) = 0 10 (a) (i) Plan A: $90
n = 3 or n = –15 (invalid) [2] Plan B: $80
(d) Perimeter: 8, 14, 20, 26… Plan C: $25 [1]
T = a + (n – 1)d
n
Answers 163
Answers.indd 163 15/03/2022 11:08 AM