Form 5 Chapter 12 (Progressions) - Spm Practice Worksheet With Answers Page 2
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1
2
3
4
57 (a) If 3, k, 48 are in an arithmetic progression,
11 (a)
T
= 10
3
then
2
… 1
ar
= 10
k – 3 = 48 – k
2k = 51
T
+ T
= 15
3
4
k = 25.5
10 + T
= 15
4
T
= 5
(b) If 3, k, 48 are in a geometric progression, then
4
3
… 2
ar
= 5
k
48
=
3
k
2
3
ar
5
:
=
2
k
= 144
2
ar
10
1
k = 12
1
r =
2
8 (a) The common difference of the arithmetic
From 1 :
progression 2, 5, 8, … is 5 – 2 = 3.
2
1
(b) The sum of all the terms from the 4th term to
a
= 10
2
the 23rd term
= S
– S
1
23
3
a
= 10
23
3
4
=
[2(2) + (23 – 1)(3)] –
[2(2) + (3 – 1)(3)]
2
2
a = 40
= 805 –15
= 40
= 790
(b) S
= 80
∞
1
1 –
2
9 (a) The common ratio of the geometric
6
progression 2, 6, 18, … is
= 3.
2
12 (a) 100y, 50y, 25y, … is a geometric
progression.
(b)
S
= 6560
n
n
a(r
– 1)
T
50y
= 6560
1
2
(b)
=
=
r –1
T
100y
2
1
n
2(3
– 1)
= 6560
T
25y
3 – 1
1
3
=
=
T
50y
2
n
3
– 1 = 6560
2
n
3
= 6561
T
T
1
2
3
Since
=
=
(a constant), the
n
8
3
= 3
T
T
2
1
2
∴ n = 8
progression is a geometric progression.
10
T
= 3k +1
8
13
T
– T
= T
– T
a + 7d = 3k + 1
2
1
3
2
9 – (t – 2) = 3t – 9
a + 7(4) = 3k + 1
9 – t + 2 = 3t – 9
… 1
a = 3k – 27
20 = 4t
t = 5
S
= 13k + 6
8
8
∴ d = 9 – (5 – 2)
(2a + 7d ) = 13k + 6
2
= 6
8a + 28d = 13k + 6
a
= 50
8a + 28(4) = 13k + 6
14 S
=
= 125
∞
1 – r
8a = 13k –106 … 2
3
1 –
5
Substituting 1 into 2 ,
8(3k – 27) = 13k –106
–12
1
15 r =
= –
24k – 216 = 13k –106
24
2
11k = 110
1
× 6 = –3
h =
k = 10
2
41
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