Form 5 Chapter 12 (Progressions) - Spm Practice Worksheet With Answers
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5Form 5: Chapter 12 (Progressions)
SPM Practice
Fully-Worked Solutions
Paper 1
1 (a) k + 1, k + 5, 2k + 6, … → A.P.
4 The arithmetic progression 5, 9, 13, … has a
common difference of 4.
d = T
– T
= k + 5 – (k + 1) = 4
2
1
d = T
– T
= 2k + 6 – (k + 5) = k + 1
3
2
These three terms
Since common difference is always the
are not the first three
S
= 57
same, k + 1 = 4 ⇒ k = 3
3
terms but any three
3
[2a + (3 – 1)(4)] = 57
consecutive terms with
2
(b) When k = 3, we have 4, 8, 12, …
a common difference
8
of 4. Therefore, a new
S
=
[2(4) + 7(4)] = 144
3
8
2
value of a (first term)
(2a + 8) = 57
2
has to be determined.
3a + 12 = 57
a = 15
2 (a)
T
= 24
4
3
ar
= 24
Hence, the three consecutive terms which sum up
3
81r
= 24
to 57 are 15, 19 and 23.
24
3
r
=
81
+4 +4
8
=
27
2
5 Volumes of water in litres:
r =
3
410, 425, 440, …
a
Volume of water at the end of the 8th day
(b) S
=
∞
1 – r
= T
8
= 81
= a + 7d
= 410 + 7(15)
2
1 –
= 515 litres
3
= 243
6 0.848484 …
= 0.84 + 0.0084 + 0.000084 + …
9
3 Since k, 3,
, m are four consecutive terms
0.84
k
=
1 – 0.01
of a geometric progression,
0.84
3
= m
=
a
S
=
0.99
k
∞
1 – r
9
84
=
k
99
3
mk
28
=
=
k
9
33
2
mk
= 27
27
m =
2
k
40
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