Chapter 4 Exercise 4.1
Q. 1.
(i)
Q 6.
Q.

(ii) ÷
(ii) 94 ÷ 4 = 23 R 2 ⇒ hexagon
(iii) ∠
(iii) 100 ÷ 4 = 25 R 0 ⇒ yellow
(iv) ¾
(iv) If n is even, then Tn will always be
a hexagon-shaped tile, whereas if
n is odd, Tn will always be a
triangle-shaped tile.
(v)
Q. 2.
(i) red, red, blue
Q
Q. 7.
(ii) blue, blue, red
(iv) brown, blue, green
(a)
(i) red, blue, red
(ii) red, blue, red
Q. 8.
Q
(iii) red, red, red
(iv) red, blue, blue
Q. 4.
Q. 5.
(i) The pattern is the pattern of primes:
2, 3, 5, 7, 11, 13, …
(ii)
(v) red, yellow, red
(b)
red, red, red, red, red, blue, red, red, red
The pattern has increasing numbers of
red squares between two individual blue
squares. So, the next run of red squares
will have 5 red squares in it, and after that
6, etc. …
(iii) blue, red, green
Q. 3.
(i) 15 ÷ 4 = 3 R 3 ⇒ triangle
(i) 50th: blue
150th: blue
100th: blue
(ii) 50th: blue
150th: blue
100th: blue
(iii) 50th: red
150th: red
100th: blue
(i) The pattern is 2, 3, 5, 7, 2, 3, 5, 7,
2, …
(iv) 50th: green
150th: green
100th: green
(ii)
(v) 50th: yellow
150th: brown
100th: green
OR
(i) 50th: 50 ÷ 2 = 25 R 0 ⇒ hexagon
(ii) 93rd: 93 ÷ 2 = 46 R 1 ⇒ triangle
Exercise 4.2
E
(iii) nth: even ÷ 2 = some whole
number R 0 ⇒ hexagon
Q. 1.
Q
(i) 40 ÷ 3 = 13 R 1 ⇒ triangle
(ii)
(ii) 30 ÷ 3 = 10 R 0 ⇒ square
(iii) 3n ÷ 3 = n R 0 ⇒ square
Reason: 3n ÷ 3 has remainder 0
(i.e. 3n is a multiple of 3)
(i)
T4 =
T5 =
T1
T2
T3
T4
T5
6
9
12
15
18
(iii) Arithmetic sequence, first term 6,
common difference 3.
(iv) T8 = 6 + 7(3) = 27
(v) Tn = 6 + (n − 1)(3) = 3n + 3
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
1
Q. 2.
(i)
(ii)
T1
T2
T3
T4
T5
4
7
10
13
16
(vii) Yes. First term = 12.
Common difference = 3
1 1 1 1 1 1
(viii) No. __ − __ ≠ __ − __ ≠ __ − __
3 2 4 3 5 4
1 1
≠ __ − __ (no common difference)
6 5
(ix) Yes. First term = −5.
Common difference = 4
(iii) Arithmetic sequence, first term 4,
common difference 3
(x) Yes. First term = 20.
Common difference = 1
(iv) T9 = 4 + 8(3) = 28
(v) Tn = 4 + (n − 1)(3) = 3n + 1
Q. 3.
Q. 6. (a)
Q
(i)
(i) 2
(vi) −5
(ii) 5
(vii) 5.5
(iii) 19
(viii) 1
(iv) 100
(ii)
T1
T2
6
8
T3
10
T4
T5
12
14
(v) 13
(b)
(iii) Arithmetic sequence, first term 6,
common difference 2
(ii) 2
(iii) −3
(iv) T7 = 6 + 6(2) = 18
(iv) −10
(v) Tn = 6 + (n − 1)(2) = 2n + 4
(v) 7
(vi) n − 1
Q. 4.
(i) 4
(c)
(i)
(i) 14, 18, 22
(ii) 13, 15, 17
(iii) 10, 7, 4
(ii)
T1
T2
T3
T4
T5
6
9
12
15
18
(iii) Arithmetic sequence, first term 6,
common difference 3
(iv) 70, 60, 50
(v) 34, 41, 48
Q
Q. 7. (a)
(ii) −10
(iv) T7 = 6 + 6(3) = 24
(iv)
(i) Yes. First term = 3.
Common difference = 2
(v)
(ii) Yes. First term = 2.
Common difference = 2
(b)
(vi) 1, 3, 5
(vii) 7, 7.5, 8
3
1
(viii) 1 __, 2, 2 __
4
4
(ix) 39, 28, 17
2 5
(x) __, __, 1
3 6
1
(vi) −1__
2
3
(vii) 2 __
4
(viii) 4.4
1
4
(ix) ___
10
1
1
−2 __
(x) − __
2
8
1
1
−12, −15, −18 (vi) 12 __, 11, 9 __
2
2
3
1
__
80, 70, 60
(vii) 19 , 22, 24 __
4
4
1
(v) 5, 2 __, 0
2
3 2 1
(ix) ___, __, __
10 5 2
5 1 3
(x) __, __, __
8 2 8
(i)
3
(iv) 40
(ii)
1
(v) −11
(iv) 4, 8, 12
Q
Q. 8. (a)
(iii) −5
2
(vii) 0.5
1
(viii) __
4
(ix) −11
1
(x) __
6
(iii) 9.2, 10.8, 12.4 (viii) 3.4, 7.8, 12.2
(iv) No. 1 − 1 ≠ 2 − 1 = 3 − 2 ≠ 5 − 3
≠ 8 − 5 (no common difference)
(vi) Yes. First term = 2.
Common difference = 0.5
(i)
(ii)
(iii) No. 2 − 1 ≠ 4 − 2 ≠ 8 − 4 ≠
16 − 8 (no common difference)
(v) Yes. First term = 5.
Common difference = 5
(vi) 2
(iii) 1.6
(v) Tn = 6 + (n − 1)(3) = 3n + 3
Q. 5.
(i) −3
(ix) 72
1
(x) __
6
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
(b)
(i) 4
(iv) −5
(ii) 5
(v) 2
(iv)
(iii) 4
(c)
10
(i) 15
(iv) 25
(ii) 16
(v) −5
8
6
(iii) 7
Q. 9. (a)
(i) 3, 5, 7, 9
4
(ii) 2, 5, 8, 11
2
(iii) 4, 7, 12, 19
0
0
(iv) 10, 8, 6, 4
1 1 1
(v) 1, __, __, __
2 3 4
1
2
3
4
(v)
(b) (i), (ii) and (iv) are arithmetic.
(i)
1
0.8
10
9
8
0.6
7
6
5
0.4
4
3
2
0.2
1
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
0
0
1
2
3
4
(ii)
Q. 10. (a)
Q
11
(i) 5, 9, 13, 17
10
(ii) 10, 7, 4, 1
9
8
(iii) 5, 11, 21, 35
7
6
(iv) 7, 4, 1, −2
5
4
(v) 4, 16, 64, 256
3
2
(b) (i), (ii) and (iv) are arithmetic.
1
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
(i)
5
(iii)
16
14
20
12
18
16
10
14
8
12
6
10
4
8
2
6
0
0
1
2
3
4
4
2
0
0
1
2
3
4
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
3
(ii)
Q. 11. (a)
Q
Sequence C
(b)
Sequence A: 2, 4, 8
Sequence B: 1, 4, 9
Sequence C: 5, 9, 13
8
(c)
4
6
Q. 12. (a)
Q
4
(b)
Sequence A: 3, 9, 27
Sequence B: 2, 5, 10
Sequence C: 3, 5, 7
(c)
2
10
2
0
0
1
2
3
4
Q. 13.
Q
(iii)
Sequence C
(i) dA = 6, dB = 2, dC = 3
(ii) mA = 6, mB = 2, mC = 3
(iii) Slope = common difference
35
Q
Q. 14. 3x − 1 − (x + 3) = 4x − 1 − (3x − 1)
30
25
2x − 4 = x
20
x=4
15
Q
Q. 15. 3x − (x + 1) = 2x + 8 − 3x
10
2x − 1 = −x + 8
5
3x = 9
0
0
1
2
3
4
x=3
(iv)
Q
Q. 16.
(i) 20 − (3x + 2) = 2x + 3 − 20
−3x + 18 = 2x − 17
6
35 = 5x
x=7
4
(ii) 20 − (3(7) + 2) = 20 − 23 = −3
2
(iii) T4 = 23 + 3(−3) = 23 − 9 = 14
0
0
1
2
3
Q
Q. 17.
4
(i) 2x + 1 − (3x − 2) = 18 − x − (2x + 1)
−x + 3 = −3x + 17
–2
2x = 14
(v)
x=7
(ii) Sequence is 19, 15, 11.
250
∴ Common difference = −4
200
(iii) T4 = 19 + 3(−4) = 19 − 12 = 7
T5 = 19 + 4(−4) = 19 − 16 = 3
150
100
50
0
0
4
1
2
3
4
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
Q
Q. 8.
Exercise 4.3
Q. 1.
(a) (i) 5
(ii) T55 = 4(55) + 1 = 221
(vi) −12
(ii) 4
(vii) 43
(iii) 1
(viii) −16
(iv) 13
(ix) 75
(v) 59
(x) −20
(b) (i) 2
Q. 9.
Q
(vii) −3
(iii) 4
(viii) −4
(iv) 7
(ix) 9
(v) −2
(x) 3
3n = 105
n = 35
}
Q. 10. a = 91 Tn = 91 + (n − 1)(−2)
Q
d = −2
= −2n + 93
−2n + 93 > 0
93 > 2n
2n < 93
(c) (i) 5 + (n − 1)(2) = 2n + 3
n < 46.5
(ii) 4 + (n − 1)(3) = 3n + 1
∴ Ans = 46
(iii) 1 + (n − 1)(4) = 4n − 3
(iv) 13 + (n − 1)(7) = 7n + 6
(v) 59 + (n − 1)(−2) = −2n + 61
(vi) −12 + (n − 1)(5) = 5n − 17
(vii) 43 + (n − 1)(−3) = −3n + 46
(viii) −16 + (n − 1)(−4) = −4n − 12
(ix) 75 + (n − 1)(9) = 9n + 66
(x) −20 + (n − 1)(3) = 3n − 23
Q. 2.
}
(i) a = −11 Tn = −11 + (n − 1)(−4)
d = −4
= −4n − 7
}
(i) a = 10 Tn = 10 + (n − 1)(12)
d = 12
= 12n − 2
}
}
}
Q.
Q 12. a = 100 Tn = 100 + (n − 1)(−3)
d = −3
= −3n + 103
∴ −3n + 103 > 0
3n < 103
n < 34.3
∴d=8
(i) a = 7 Tn = 7 + (n − 1)(6)
d=6
= 6n + 1
Q. 14. Tn = a + (n − 1)(d) = a + 15(n − 1) =
Q
15n − 15 + a
}
(i) a = 3 Tn = 3 + (n − 1)(8)
d=8
= 8n − 5
}
∴ Answer = 21
144 = 18d
T51 = 248 = 15(51) − 15 + a = 750 + a
∴ 750 + a = 248
(ii) T96 = 8(96) − 5 = 763
Q. 7.
n < 21.75
T19 = 150 = 6 + 18d
(ii) T33 = 6(33) + 1 = 197
Q. 6.
4n < 87
Q
Q. 13. Tn = a + (n − 1)(d) = 6 + (n − 1)(d)
(i) a = 0 Tn = 0 + (n − 1)(7)
d=7
= 7n − 7
(ii) T85 = 7(85) − 7 = 588
Q. 5.
4n + 13 < 100
∴ Answer = 34
(ii) T64 = 12(64) − 2 = 766
Q. 4.
}
Q
Q. 11. a = 17 Tn = 17 + (n − 1)(4)
d=4
= 4n + 13
103 > 3n
(ii) T55 = −4(55) − 7 = −227
Q. 3.
}
(i) a = 14 Tn = 14 + (n − 1)(3)
d=3
= 3n + 11
∴ 3n + 11 = 116
(vi) 5
(ii) 3
}
(i) a = 5 Tn = 5 + (n − 1)(4)
d=4
= 4n + 1
∴ a = −502
(i) a = 31 Tn = 31 + (n − 1)(–6)
d = –6
= –6n + 37
(ii) T21 = −6(21) + 37 = −89
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
5
Q. 15. Tn = a + (n − 1)(d) = a + (n − 1)(5)
}
(i) a = 30 Tn = 30 + (n − 1)(5)
d=5
= 5n + 25
Q
Q. 17.
T40 = 480 = a + 39(5)
Q. 16.
480 = a + 195
T1 ⇒ 2011
∴ a = 285
T11 ⇒ 2021
T14 ⇒ 2024
(i)
A
B
Row 1
Row 2
7
6
Row 3
Row 4
13
12
E
2
3
4
19
18
24
Row 9
= 95
(ii) 5n + 25 = 75
9
5n = 50
10
n = 10
11
15
⇒ 2020
16
(iii)
17
20
25
T14 = 5(14) + 25
5
14
Row 7
Row 8
D
8
Row 5
Row 6
C
21
22
No. of trees planted
1
100
23
26
1
120
27
(ii) Row 100 = T50
}
a = 7 Tn = 7 + (n − 1)(6)
d=6
= 6n + 1
∴ T50 = 301
(iii) Row 1 = T1
28
80
60
40
20
Year no. (15 = 2025, 16 = 2026, etc.)
0
0
Q. 18.
Q
Row 3 = T2
2
4
(i)
6
8
10
14
16
18
20
22
Day
Number of hours
wearing lenses
1
2
2
2.25
a=4
3
2.5
d=6
4
2.75
5
3
Row 5 = T3
Row n = T
(
n+1
______
2
Row 99 = T50
)
Tn = 4 + (n − 1)(6)
= 6n − 2
T50 = 298
(iv) Column A: Tn = 6n + 1
6n + 1 = 2011
6n = 2010
n = 335
∴ Row 335 × 2 = Row 670 of
Column A
6
12
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
}
(ii) a = 2
Tn = 2 + (n − 1)(0.25)
d = 0.25
= 0.25n + 1.75
0.25n + 1.75 = 14
0.25n = 12.25
n = 49
Answer: On day 49
24
Q. 19.
(i)
Floor number
No. of rooms
6
540
7
630
8
720
9
810
10
900
(iii) a = 4
}
T20 = 90(20) = 1,800
Q. 4.
Q
Q. 5.
Q
Q. 6.
Q
Exercise 4.4
Q. 2.
S30 =
(ii) a = 3
d=3
30
___
2
(2(1) + 79(1)) = 3240
a=1
d=2
n = 30
a=2
d=2
80 = 2n
n = 40
∴ 40 terms
40
S40 = ___ (2(2) + 39(2))
2
= 1640
Q. 7.
Q
(i) a = 1
(i) a = 2
S20 =
n = 30
(i) a = −11
d = 12
n = 10
10
___
S10 =
(2(−11) + 9(12)) = 430
2
(ii) a = −8
d=6
n = 10
10
___
(2(−8) + 9(6)) = 190
S10 =
2
d=6
n = 10
10
S10 = ___ (2(1) + 9(6)) = 280
2
(ii) a = 1
d=6
n = 20
20
S20 = ___ (2(1) + 19(6)) = 1,160
2
(iii) 1,160 − 280 = 880
Q. 8.
Q
(2(4) + 29(3)) = 1,425
d=5
= 2n
n = 30
30
S30 = ___ (2(3) + 29(5)) = 2,265
2
(iii) a = −5
d=7
n = 30
30
___
S30 =
(2(−5) + 29(7)) = 2,895
2
(iv) a = 35
d = −2
n = 30
30
___
S30 =
(2(35) + 29(−2)) = 180
2
Q. 3.
2
n = 80
Tn = 2 + (n − 1)(2)
(i) a = 11
d = 11
n = 20
20
___
S20 =
(2(11) + 19(11)) = 2,310
2
(ii) a = 5
d=2
n = 20
20
___
S20 =
(2(5) + 19(2)) = 480
2
(iii) a = 43
d = −3
n = 20
20
___
S20 =
(2(43) + 19(−3)) = 290
2
(iv) a = 1
d=4
n = 20
20
___
S20 =
(2(1) + 19(4)) = 780
2
(v) a = 13
d=7
n = 20
20
___
S20 =
(2(13) + 19(7)) = 1,590
2
(i) a = 4
d=1
80
___
30
S30 = ___ (2(1) + 29(2)) = 900
2
101 = n
Q. 1.
a=1
S80 =
Answer = 1,800
(iii) 9,090 = 90n
n = 10
(2(4) + 9(−6)) = −230
2
(iv) a = −8
d = −2
n = 10
10
___
S10 =
(2(−8) + 9(−2)) = −170
2
(v) a = −4
d=2
n = 10
10
___
S10 =
(2(−4) + 9(2)) = 50
2
S10 =
(ii) a = 90 Tn = 90 + (n − 1)(90)
d = 90
= 90n
d = −6
10
___
(ii) a = 2
S40 =
d=8
20
___
2
n = 20
(2(2) + 19(8)) = 1,560
d=8
n = 40
40
___
(2(2) + 39(8)) = 6,320
2
(iii) 6,320 − 1,560 = 4,760
Q. 9.
Q
(i) a = −1
d=8
n = 10
10
___
(2(−1) + 9(8)) = 350
S10 =
2
(ii) a = −1
d=8
n = 20
20
___
S20 =
(2(−1) + 19(8)) = 1,500
2
(iii) 1,500 − 350 = 1,150
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
7
Q. 10.
(i) a = 2
d=3
Q
Q. 13.
(i) a = 2
Tn = 2 + (n − 1)(3)
Tn = 2 + (n − 1)(0.25)
= 3n − 1
= 0.25n + 1.75
65 = 3n − 1
T30 = 0.25 (30) + 1.75
66 = 3n
= 9.25 km
22 = n
22
S22 = ___ (2(2) + 21(3))
2
= 737
(ii) Tn = 0.25n + 1.75
10 = 0.25n + 1.75
8.25 = 0.25n
319
(ii) a = 11
d = 11 n = ____ = 29
11
29
S29 = ___ (2(11) + 28(11))
2
= 4,785
(iii) a = 88
∴ 33rd week
(iii) a = 2 × 7 = 14
d = 0.25 × 7 = 1.75
d = −2
n = 49
49
S49 = ___ (2(14) + 48(1.75))
2
= 2,744 km
Tn = 88 + (n − 1)(−2)
= −2n + 90
8 = −2n + 90
Q. 14.
Q
(i) a = 20
2n = 82
Let −2n + 22 = 0
22 = 2n
n = 11
d = −4
Answer: 11 hours
Tn = 55 + (n − 1)(−4)
(ii) a = 20
d = −2
n = 11
11
S11 = ___ (2(20) + 10(−2)) = 110 cm
2
= −4n + 59
11 = −4n + 59
Q
Q. 15.
4n = 48
(i) Row
number
n = 12
12
S12 = ___ (2(55) + 11(−4))
2
= 396
99
(v) a = 3
d=3
n = ___ = 33
3
33
S33 = ___ (2(3) + 32(3)) = 1,683
2
Q. 11.
(i) a = 5
No. of red
seats
n = 5 × 12 = 60
Tn = 100 + (n − 1)(5) = 5n + 95
T60 = 5(60) + 95 = 395€
(ii) S60 =
8
60
___
2
5
7
9
11
2
3
4
5
6
7
50 ÷ 10 = 5
(ii) T55 = 4(55) + 1 = 221
50
(iii) S50 = ___ (2(5) + 49(4)) = 5,150
2
d=5
3
51 − 1 = 50
d=4
(i) a = 100
1
⇒ Every 10 rows ⇒ 5 more red seats
per row
5 × 5 = 25
Tn = 5 + (n − 1)(4) = 4n + 1
Q. 12.
d = −2
Tn = 20 + (n − 1)(−2) = −2n + 22
n = 41
41
S41 = ___ (2(88) + 40(−2))
2
= 1,968
(iv) a = 55
d = 0.25
(2(100) + 59(5)) = 14,850€
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
25 + 2 = 27 red seats
(ii)
Row number
8
No. of red seats
5
98 − 8 = 90
90 ÷ 10 = 9
9 × 5 = 45
45 + 5 = 50 red seats
(iii)
Row pairs
1
2
(v) The dots on the base form the
sequence
3
Rows Rows Rows
1+2
3+4
5+6
No. of red
seats
4
6
1, 2, 3, 4, …
L1 = 1, L2 = 2, L3 = 3, L4 = 4, …
8
∴ Ln = n
n = 100 ÷ 2 = 50
(vi) The dots on the height form the
sequence
a=4
d=2
1, 2, 3, 4, …
50
___
(2(4) + 49(2))
2
= 2,650 red seats
S50 =
H1 = 1, H2 = 2, H3 = 3, …
∴ Hn = n
(vii) Number of dots in the nth rectangle:
Tn = (n)(n)
Exercise 4.5
= n2
Q. 1.
(i)
Q. 3.
Q
(ii)
T1
T2
T3
T4
T5
1
3
6
10
15
(iii) 2
3
+2
6
+3
+1
(ii)
+5
+4
+1
15
10
(i)
+1
(iii)
∴ Second difference is a constant.
T1
T2
T3
T4
T5
1
9
16
25
36
9
16
25
4
+5
⇒ Quadratic
+11
+2
+2
∴ Second difference is a constant.
(iv) T6 = T5 + 6 = 21
⇒ Quadratic
T7 = T6 + 7 = 28
Q. 2.
+9
+7
+2
36
(iv) T6 = 36 + 13 = 49
(i)
T7 = 49 + 15 = 64
(v) Tn = n2
(ii)
T1
T2
T3
T4
T5
1
4
9
16
25
(iii) 1
4
+3
9
+5
+2
+2
+9
+2
Second difference is a constant.
⇒ Quadratic
(iv) T6 = 36
T7 = 36 + 13 = 49
T8 = 49 + 15 = 64
(i) -1
9
2
+7
+3
25
16
+7
Q. 4.
Q
+4
20
+11
+4
Quadratic (constant second differences)
(ii) Exponential (each term is double the
previous term)
(iii) 1
11
5
+6
+4
+2
19
+8
+2
Quadratic (constant second
differences)
Active
i Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
i
9
(iv) 5
45
20
+25
+15
+10
(c)
80
(i) 56, 78, 104
(ii) 15, 21, 28
+35
(iii) 79, 112, 151
+10
(iv) 67, 93, 123
Quadratic (constant second
differences)
(v) 5
15
10
+5
+5
(v) 95, 135, 183
(vi) 8, 2, −6
20
(vii) −25, −43, −65
+5
(viii) 7, 16, 28
Linear (constant first differences)
(vi) Exponential (each term is triple the
previous term)
(vii) 3
13
6
+7
+3
+4
(ix) 10, 19, 31
(a)
Q.
Q 6.
24
(ii) Triples
(iii) Triples
+11
(iv) Doubles
+4
(v) Triples
Quadratic (constant second
differences)
(viii) 6
10
8
+2
+2
(b)
12
(iii) 2673, 8019, 24057
+2
(ix) Exponential (each term is double the
previous term)
(iv) 416, 832, 1664
108
(v) −1215, −3645, −10935
How
Amount (mg)
150
1
150 (0.8) = 120
∴ Exponential (each term is triple
the previous term)
2
150 (0.8)2 = 96
3
150 (0.8)3 = 76.8
(i) 8
(vi) 8
4
150 (0.8)4 = 61.44
(ii) 1
(vii) 5
5
150 (0.8)5 = 49.152
6
150 (0.8)6 = 39.3216
(iii) 7
(viii) 1
(iv) 3
x3
(ix) 10
(v) 15
(b)
(i)
Q 7.
Q.
0
x3
x3
Q. 5. (a)
36
12
(i) 256, 512, 1024
(ii) 1458, 4374, 13122
Linear (constant first differences)
(x) 4
(i) Doubles
(ii)
Amount (micrograms)
140
1st
2nd
(i) 6, 10, 14
(i) 4
(ii) 2, 3, 4
(ii) 1
(iii) 9, 15, 21
(iii) 6
(iv) 10, 14, 18
(iv) 4
120
100
80
60
40
20
(v) 8, 16, 24
(v) 8
(vi) 4, 2, 0, −2
(vi) −2
(vii) 2, −2, −6, −10
(vii) −4
(viii) −3, 0, 3
(viii) 3
(ix) −6, −3, 0, 3
10
(ix) 3
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
Hour
0
0 0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
(iii) No. (The following hour there will
always be 80% of the amount
(non-zero) the previous hour.)
6
Q. 8.
Revision Exercises
R
(i) 53 = 125
63 = 216
Q. 1.
Q
73 = 343
(ii)
Terms No. of
tubes
1
1
2
8
3
27
4
64
5
125
6
216
1st
Diff.
+7
+19
+37
+61
+91
2nd 3rd
Diff. Diff.
+12
+18
+24
+30
+6
+6
+6
(ii) T1 = 1
(iii) Since the third difference is a constant.
Q. 9.
(i)
T2 = 5
(i) Terms Number 1st
2nd
3rd
Diff. Diff. Diff.
1
2
3
4
5
6
1
7
19
41
77
131
+6
+12
+22
+36
+54
T3 = 9
T4 = 13
+6
+10
+14
+18
Tn = 1 + (n − 1)(4) = 4n − 3
+4
+4
+4
(iii) T50 = 4(50) − 3 = 197
(iv) 237 = 4n − 3
240 = 4n
(ii) Since the third difference is a constant.
n = 60
(iii) T7 = 131 + 54 + 18 + 4 = 207
Q. 10.
Answer: 60th term
(i)
Q. 2.
Q
(i)
(ii) T1 = 3
T2 = 5
T3 = 7
T4 = 9
(ii) T5: 45
Tn = 3 + (n − 1)(2) = 2n + 1
T6: 66
(iii) T50 = 2(50) + 1 = 101
(iii) T1 = 1
(iv) 200 = 2n + 1
T2 = 6
2n = 199
T3 = 15
n = 99.5
T4 = 15 + 13 = 28
Answer = 99
T5 = 28 + 17 = 45
Q. 3.
Q
T6 = 45 + 21 = 66
1st Diff
2nd Diff
+5
+4
+9
+4
+0
+13
+4
+0
+17
+4
+0
(i)
3rd Diff
(ii)
+21
(iv) Quadratic (second difference is a
constant)
T1
T2
T3
T4
T5
5
9
13
17
21
(iii) Linear (constant first difference)
(iv) Tn = 5 + (n − 1)(4) = 4n + 1
(v) T8 = 4(8) + 1 = 33
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
11
Q. 4.
(iii)
(i) a = T1 = 5(1) + 1 = 6
(ii) d = 5
Tn
(iii) 5n + 1 = 156
1st Diff.
5n = 155
(i)
1
1
7
21
8
28
2
3
4
5
3
12
29
54
87
9
17
25
33
8
8
8
2nd Diff.
n = 31
12
(iv) S12 = ___ (2(6) + 11(5)) = 402
2
Q. 5.
1
35
56
35
70
Q. 9.
21
56
7
28
Linear ⇒ B (straight line)
1
8
(i) Quadratic ⇒ A (not straight line)
(ii)
1
(ii) Linear : a = 1
Quadratic
3
9
19
33
Linear
3
5
7
9
T1
T2
T3
T4
d=1
Tn = 1 + (n − 1)(1) = n
(iii)
(iii) Quadratic:
1
3
6
+2
+1
Q
Q. 10.
T5
T6
(ii)
(iii)
Q
Q. 11. (a)
(i)
(ii)
T1
2
(iii) 2
T2
5
T3
9
5
+3
T4
14
9
20
+6
+5
+1
T7 = 27 + 8 = 35
+9
54
29
+25
+17
+8
T4
1
4
13
27
1
__
4
1
__
4
3
37
9
+ ___ + ___ = ___
16 64 64
Pattern
1
2
3
4
5
No. of tiles
21
22
45
57
69
}
= 12n + 9
(c) T10 = 129
(d) 399 = 12n + 9
12n = 390
+1
(iv) T6 = 20 + 7 = 27
12
T3
d = 12
Quadratic (constant second difference)
(i) 3
T2
(b) Linear: a = 21 Tn = 21 + (n − 1)(12)
T5
20
14
+4
+1
T1
Answer: 27
∴ 10 minutes + 84 minutes
= 94 minutes.
+8
Quadratic (second difference is
constant)
(ii) T5 = 87
12
13
(i) 0.5 × 10 + 3 = 5 + 3 = 8 litres
42 ÷ 0.5 = 84 minutes
Q. 8.
11
(i)
(ii) 50 − 8 = 42 litres
Q. 7.
Linear
Answer = 73 − 13 = 60
+1
Second difference is constant.
Q. 6.
73
+5
+4
+1
51
15
10
+3
Quadratic
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
n = 32.5
Answer = 32nd pattern
n
(e) Sn = __ (2(21) + (n − 1)(12))
2
n
= __ (42 + 12n − 12)
2
n
= __ (30 + 12n)
2
= n(15 + 6n)
= 6n2 + 15n
(f) 6n2 + 15n = 399
6n2 + 15n − 399 = 0
2n2 + 5n − 133 = 0
−5 ± 33
n = ________ (Quadratic formula: see Active Maths 3, Book 1, Chapter 7, page 149)
4
−38 ___
28
____
n=
or
4
4
−38
∴n=7
Reject n = ____
4
since n > 0, n ∈ N
∴ 7 patterns
Q. 12. (a)
Day 1
Day 2
Day 3
Day 4
Day 5
Day 6
Day 7
Plant 1
16
20
24
28
32
36
40
Plant 2
24
27.5
31
34.5
38
41.5
45
(All heights in cm)
(b) Plant 1: height = 16 + (n − 1)(4)
height = 4n + 12
Plant 2: height = 24 + (n − 1)(3.5)
height = 3.5n + 20.5
(n = no. of days)
(c)
Height (cm)
120
100
80
(17,80)
60
40
Plant 2
20
Plant 1
Day
0
0
(d) (i)
2
4
6
8
10 12 14 16 18 20 22 24 26 28
(17, 80)
(ii) At day 17, both plants will be 80 cm tall.
(e) 4(17) + 12 = 80
3.5(17) + 20.5 = 80
68 + 12 = 80
59.5 + 20.5 = 80
80 = 80
80 = 80
TRUE
TRUE
∴ Both formulae tested + verified.
(f) Any slight errors in constructing and/or reading the graphs of each plant’s growth are
excluded.
(g) The growth rate of a plant for the first week is unlikely to be the same after, say, two or three
weeks. It is likely to decrease eventually. (Extrapolation outside of the range of values over
which data is collected has limitations.)
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions
13
Q. 13. (a) T1 = 38 − 4(1) = 34
T2 = 38 − 4(2) = 30
T3 = 38 − 4(3) = 26
T4 = 38 − 4(4) = 22
Ans: 34, 30, 26
(b) 38 − 4n < 0
4n > 38
n > 9.5
Ans: T10
(c)
n
Sn = __ [2a + (n − 1)d]
2
15
S15 = ___ [2(34) + (15 − 1)(−4)]
2
15
___
=
[68 − 56]
2
= 90
Ans = 90
n
__
[2(34) + (n − 1)(−4)] = 0
2
n
__
[68 − 4n + 4] = 0
2
n
__
[72 − 4n] = 0
2
n [36 − 2n] = 0
36 − 2n = 0 ⇒ 2n = 36 ⇒ n = 18
Ans = 18
14
Active Maths 3 Book 1 – Strands 1–5 – Ch 4 Solutions