chapter 1  progressions
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8/3/2019 Chapter 1  Progressions
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Progressions1CHAPTER
(a) 2, 5, 8, 11, 14
T1 T
2 T
3T
4 T
5
1.1 Arithmetic Progression
Example
A Determine whether each of the following is an arithmetic progression.
T2
T1
= 5 2 = 3
T3
T2
= 8 5 = 3
T4
T3
= 11 8 = 3
T5
T4
= 14 11 = 3
d= 3 is a constant.
Therefore 2, 5, 8, 11, 14 is an arithmetic
progression.
(b) 2, 0, 2, 5
T1 T
2 T
3T
4
T2
T1
= 0 2 = 2
T3
T2
= 2 0 = 2
T4
T3
= 5 (2) = 3
dis not a constant.
Therefore 2, 0, 2, 5 is not an arithmetic
progression.
1. 4, 7, 10, 13, .....
T2 T1 = 7 4= 3
T3
T2
= 10 7
= 3
T4
T3
= 13 10
= 3
d= 3
Therefore, 4, 7, 10, 13 is an arithmetic
progression.
2. 6, 11, 16, 21, .....
T2 T1 = 11 6= 5
T3
T2
= 16 11
= 5
T4
T3
= 21 16
= 5
d= 5
Therefore, 6, 11, 16, 21 is an arithmetic
progression.
3. 8, 11, 14, 18, ..... 4. 8, 5, 2, 1, .....
T2
T1
= 5 8
= 3
T3
T2
= 2 5
= 3
T4
T3
= 1 2
= 3
d= 3
Therefore, 8, 5, 2, 1 is an arithmetic
progression.
T2
T1
= 11 8
= 3
T3
T2
= 14 11
= 3
T4
T3
= 18 14
= 4
dis not a constant.
Therefore, 8, 11, 14, 18 is not an arithmetic
progression.

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B Find the sixth term of the following arithmetic progressions.
Example
6, 13, 20, .....
a = 6, d= 13 6
= 7
Tn
= a + (n 1)d
T6
= 6 + (6 1)7
= 41
1. 3, 7, 11, .....
a = 3, d= 11 7
= 4
Tn
= a + (n 1)d
T6
= 3 + (6 1)(4)
= 23
3. 3, 2, 7, .....
a = 3, d= 2 3
= 5
Tn
= a + (n 1)d
T6
= 3 + (6 1)(5)
= 22
2. 6, 2, 2, .....
a = 6, d= 2 6
= 4
Tn
= a + (n 1)d
T6
= 6 + (6 1)(4)
= 14
5. 5, 9, 13, 17, .....
a = 5, d= 9 (5)
= 4
Tn
= a + (n 1)d
T6
= 5 + (6 1)(4)
= 25
4. 12, 7, 2, 3, .....
a = 12, d= 7 12
= 5
Tn
= a + (n 1)d
T6
= 12 + (6 1)(5)
= 13
5. 1.1, 0.8, 0.5, 0.1, .....
T2
T1
= 0.8 1.1
= 0.3
T3
T2
= 0.5 0.8
= 0.3
T4
T3
= 0.1 0.5
= 0.4
dis not a constant.
Therefore, 1.1, 0.8, 0.5, 0.1 is not an arithmetic
progression.
6. 7, 2, 3, 8 .....
T2
T1
= 2 (7)
= 5
T3
T2
= 3 (2)
= 5
T4
T3
= 8 3
= 5
d= 5
Therefore, 7, 2, 3, 8 is an arithmetic
progression.

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C Solve the following problems.
Example
The third term and the fifth term of an arithmetic
progression are 23 and 39 respectively. Find the
first term and common difference.
Tn
= a + (n 1)d
T3
= a + (3 1)d
T3
= a + 2d
T5
= a + (5 1)d
T5
= a + 4d
a + 2d= 23 1
a + 4d= 39 2
2 1 : 2d = 16
d = 8
From 1 , a + 2d = 23
a = 23 2(8)
= 7
1. The third term and the eighth term of an arithmetic
progression are 22 and 57 respectively. Find thefi
rstterm and the common difference.
Tn
= a + (n 1)d
T3
= a + (3 1)d
T3
= a + 2d
T8
= a + (8 1)d
T8
= a + 7d
a + 2d= 22 1
a + 7d= 57 2
2 1 : 5d = 35 d = 7
From 1 , a = 22 2(7)
= 8
2.
The third term and the seventh term of an arithmeticprogression are 31 and 79 respectively. Find the
first term and the common difference.
Tn
= a + (n 1)d
T3
= a + (3 1)d
T3
= a + 2d
T7
= a + (7 1)d
T7
= a + 6d
a + 2d= 31 1
a + 6d= 79 2
2 1 : 4d = 48
d = 12
From 1 , a = 31 2(12)
= 7
3.
The fourth and the sixth terms of an arithmeticprogression are 1 and 5 respectively. Find the
first term and the common difference.
T4
= a + (4 1)d
= a + 3d
T6
= a + (6 1)d
= a + 5d
a + 3d= 1 1
a + 5d= 5 2
2 1 : 2d = 6
d = 3
From 1 , a = 1 + 9
= 10

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D Find the sum of the first 6 terms of each arithmetic progression below.
Example
3, 2, 7, ... 1. 6, 13, 20, ...
a = 3, d= 2 3
= 5
S
n=
n[2a + (n 1)d]
2
S
6=
6[3 + (6 1)(5)]
2
=
6[3 25]
2
= 3(22)
= 66
a = 6, d= 13 6
= 7
S
n=
n[2a + (n 1)d]
2
S
6=
6[2(6) + (6 1)(7)]
2
=
6[47]
2
= 3(47)
= 141
3. 8, 5, 2, ...
a = 8, d= 5 8
= 3
S
n=
n[2a + (n 1)d]
2
S6
= 6 [2(8) + (6 1)(3)]
2
=
6[1]
2
= 3(1)
= 3
2. 5, 2, 1, ...
a = 5, d= 2 5
= 3
S
n=
n[2a + (n 1)d]
2
S6
= 6 [2(5) + (6 1)(3)]
2
=
6[5]
2
= 3(5)
= 15
5. 27, 22, 17, ...4. 2, 2, 6, ...
a = 2, d = 2 (2) = 4
S6
=6
[2(2) + (6 1)4]
2
= 3[4 + 5(4)]
= 3(16)
= 48
a = 27, d= 22 27 = 5
S
6=
6[2(27) + (6 1)(5)]
2
= 3[54 + 5(5)]
= 3(29)
= 87

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E Calculate the sum of all the terms for each of the arithmetic progressions below.
Example
2, 6, 10, 14, ..., 50 1. 3, 5, 7, ..., 97
Tn
= 50
a + (n 1)d= 50
2 + (n 1)4 = 50
4n 4 = 48
4n = 52
n = 13
Tn
= 97
a + (n 1)d = 97
3 + (n 1)(2) = 97
3 + 2n 2 = 97
2n = 96
n = 48
4. The sum ofn terms of an arithmetic progression is
given by Sn
= 2n2 + 5n. Find
(a) the first term,
(b) the common difference.
(a) S1
= a
Sn
= 2n2 + 5n
S1
= 2(1)2 + 5(1)
= 7
a = 7
(b) S2
= 2(2)2 + 5(2)
= 8 + 10
= 18
T2
= 18 7
= 11
T2
= a + (2 1)d
11 = 7 + d
d= 4
or
d= T2
T1
= 11 7
= 4
Sn
= n
[a + l]
2
S13
=13
(2 + 50)
2
= 338
Sn
= n
(a + l)
2
S48
=48
(3 + 97)
2
= 2400
3. 12, 7, 2, ..., 33
Tn
= 33
a + (n 1)d = 33
12 + (n 1)(5) = 33
5n 5 = 45
5n = 50
n = 10
Sn
= n (a + l)
2
S10
=10
(12 + 33)
2
= 5(21)
= 105
2. 10, 8, 6, ..., 12
Sn
= n (a + l)
2
S12
=12
[10 + (12)]
2
= 6(2)
= 12
Tn
= 12
a + (n 1)d= 12
10 + (n 1)(2) = 12
2n + 2 = 22
2n = 24
n = 12
(b) S10
S3
=10
[2(8) + (10 1)(7)]
2
3[2(8) + (3 1)(7)]
2
= 5(79)
3(30)
2
= 395 45
= 350
5. The fourth term of an arithmetic progression is
29 and the sum of the first eight terms is 260.
Find
(a) the first term and the common difference,
(b) the sum from the fourth term to the tenth term.
(a) T4
=29 S8
= 260
a + (4 1)d= 29
a + 3d= 29 1
8[2a + (8 1)d] = 260
2
4(2a + 7d) = 260
2a + 7d = 65 2
From 1 , 2(29 3d) + 7d= 65
58 6d+ 7d= 65
d= 7 a = 29 3d
= 29 3(7)
= 8

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8. Given the first three terms of an arithmetic
progression are a, 2a 1 and 2a + 3. Find
(a) the value ofa,
(b) the sum of the next 12 terms.
9. The second term of an arithmetic progression is 2.
The sixth term is more than the sum of the first ten
terms by 75. Find
(a) the first term,
(b) the common difference,
(c) the tenth term.
6. An arithmetic progression has 12 terms. The sum
of all 12 terms is 192, and the sum of the odd terms
is 90. Find
(a) the first term and the common difference,
(b) the last term.
S12
= 192; T1
+ T3
+ T5
+ T7
+ T9
+ T11
= 90
(a) 6a + 2d+ 4d+ 6d+ 8d+ 10d = 90
6a + 30d = 90
a + 5d = 15 1
a = 2; d= 6 2 = 4
Sn 200
n[2(2) + (n 1)(4)] 200
2
n(4 + 4n 4) 200
2
n(2n) 200
n2 100
n2 100 0
(n 10)(n + 10) 0
n 10 or n 10
(not accepted)
n = 11
10 10
n 0
n 0
(a) T2
= a + d
a + d= 2 1
T
6= a + 5d; S
10=
10[2a + 9d]
2
T6
S10
= 75
a + 5d 5(2a + 9d) = 75
a + 5d 10a 45d= 75
9a 40d= 75 2From 1 , d= 2 a
9a 40(2 a) = 75
9a 80 + 40a = 75
31a = 155
a = 5
(b) d = 2 a
= 2 5 = 3
(c) T10
= a + 9d
= 5 + 9(3)
= 22
12[2a + (12 1)d] = 192
2
6(2a + 11d) = 192
2a + 11d= 32
From 1 , 2(15 5d) + 11d= 32
30 10d+ 11d= 32
d= 2
a = 15 5(2)
= 5
(b) T12
= a + (12 1)d
= 5 + 11(2)
= 5 + 22
= 27
(a) T2
T1
= T3
T2
2a 1 a = 2a + 3 2a + 1
a 1 = 4
a = 5
d= T2
T1
= a 1
= 4
(b) S15
S3
= 15 [2(5) + (15 1)(4)]2
3[2(5) + (3 1)(4)]
2
=15
(66) 3
(18) 2 2
= 495 27
= 468
7. Find the least number of terms of the arithmetic
progression 2, 6, 10, 14, ..... so that the sum of all
its terms exceeds 200.

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1.2 Geometric Progression
2. 4, 11, 17, 25, ...
T2=
11
T
14
T3=
17
T
211
T4 = 25
T
317
r is not constant.
Therefore 4, 11, 17, 25, ... is not a geometric
progression
1. 4, 8, 16, 32, ...
T2=
8= 2
T
14
T3=
16= 2
T
28
T4 = 32 = 2
T
316
r = 2
Therefore 4, 8, 16, 32, ... is a geometric
progression
4. 1, 8, 64, 512, .....
T2 = 8 = 8
T
11
T3=
64= 8
T
28
T4=
512= 8
T
364
r is not constant.
Therefore 1, 8, 64, 512, ... is not a geometric
progression
3. 16, 8, 4, 2, .....
T2 = 8 = 1
T
116 2
T3=
4=
1
T
28 2
T4=
2=
1
T
34 2
r =
1
2
Therefore 16, 8, 4, 2, ... is a geometric
progression
Example
(a) 4, 8, 16, 32
T1 T
2T
3 T
4
T2=
8= 2
T
14
T3=
16= 2
T
28
T4=
32= 2
T
316
r = 2 is a constant.
Therefore 4, 8, 16, 32, ... is a geometric progression
(b) 6, 12, 24, 48
T1 T
2T
3 T
4
T2=
12= 2
T
16
T3=
24= 2
T
212
T4=
48= 2
T
324
r is not a constant.
Therefore 6, 12, 24, 48, ... is not a geometric
progression
A Determine whether each of the following is a geometric progression.

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B Find the eighth term of each of the following geometric progressions.
Example
3, 6, 12, 24, ...
5. 5, 5 , 5 , 5 , ...
2 4 8
4. 3, 12, 48, 192, ...
a = 3, r =6
3
r= 2
Tn
= ar n 1
T8
= 3(2)8 1
= 3(2)7
= 384
1. 2, 6, 18, 54, ...
a = 2, r =6
2
r= 3
Tn
= ar n 1
T8
= (2)(3)8 1
= 2(3)7
= 4374
a = 3, r =12
3
= 4
Tn
= ar n 1
T8
= (3)(4)8 1
= 3(4)7
= 49 152
a = 5, r =
5
Tn
= ar n 1
T8
= (5)( 1 )8 1
2
= 5( 1 )
7
2
=
5
128
a = 6, r =12
6
r= 2
Tn
= ar n 1
T8
= (6)(2)8 1
= 6(2)7
= 768
52
3. 7, 21, 63, 189, ...
2. 6, 12, 24, 48, ...
a = 7, r =21
7
r= 3
Tn
= ar n 1
T8
= 7(3)8 1
= 7(3)7
= 15 309
=
1
2
7. 1,
1 , 1 ,1 , ...
3 9 27
6. 4, 6, 9,
27, ...
2
a = 4, r =6
4
=32
T8
= 4( 3 )7
2
= 68
11
32
a = 1, r = 1
3
T8
= 1( 1 )7
3
=
1
2187

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C Solve the following problems.
Example
The second term and the fourth term of a geometric
progression are 6 and 54 respectively. Find the firstterm and the common ratio (r 0).
T2
= ar2 1 = ar
T4
= ar4 1 = ar3
1. The third and the fourth term of a geometric
progression are 12 and 8 respectively. Find the first
term and the common ratio.
ar3
=8
ar2 12
r =
2
3
Given
T3
= ar2 = 12
T4
= ar3 = 8
ar2 = 12
a( 2 )2
= 12
3
4 a = 12 9 a = 27
3. In a geometric progression, the second term is 8
more than the first term and the third term is 24
more than the second term. Find
(a) the common ratio,
(b) the first term,
(c) the fifth term.
(a) T2
T1
= 8;
T3
T2
= 24
ar a = 8 a(r 1) = 8 1
ar2 ar = 24
ar(r 1) = 24 2
2 1 : r = 3
2. The third term and the sixth term of a geometric
progression are 27 and 8 respectively. Find the first
term and the common ratio. Hence, find the fifth
term.
ar5
=8
ar2
27r3=
8
27
r =
2
3
Given
T3
= ar2 = 27
T6
= ar5 = 8a( 2 )
2= 27
3
4 a = 27
9
a = 60
3
4
(b) From 1 ,
a(2) = 8
a = 4
(c) T5
= ar4
= (4)(3)4
= 324
4. In the geometric progression 8, 32, 128, ..., find the
first term which exceeds 8 200.
Tn
8 200ar n 1 8 200
(8)( 32 )
n 1 8 200
8
8(4)n 1 8 200
(4)n 1 1 025
(n 1)log10
4 log10
1 025
n 1 5
n 6
n = 7
T5
=(60 3 )( 2 )
4
4 3
= 12
5. Find the number of terms in the geometric
progression 4 096, 512, 64, ...,1
. 4 096
Tn =
1
4 096
ar n 1 =
1
4 096
(4 096)( 512 )n 1
=1
4 096 4 096
84
( 1 )n 1
=1
8 84
( 1 )n 1
=1
8 88
n 1 = 8
n = 9
Given ar = 6 and ar3 = 54,
ar3
=54
ar 6
r 2 = 9
r = 3
ar = 6
a(3) = 6
a = 2
The seventh term is the first term that exceeds
8 200.

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D Calculate the sum of the first six terms for each of the geometric progressions below.
Example
1. 3, 6, 12, ...
a = 3, r =6
3
= 2
Sn
= a(r n 1)
r 1
S6
=3[(2)6 1]
2 1
= 63
a = 2, r =6
2
= 3
Sn
= a(r n 1)
r 1
S6
=2(36 1)
3 1
= 728
3. 2, 8, 32, ...
a = 2, r =8
2
= 4
Sn
=a(r n 1)
r 1
S6
=2(46 1)
4 1
= 2 730
2. 3, 6, 12, ...
5. 2, 1,1
, ...
2
4. 3, 6, 12, ...
a = 3, r =6
3
= 2
Sn
=a(r n 1)
r 1
S6
=3[(2)6 1]
2 1
= 63
7. 12, 4, 4 , ...
3
a = 12, r =
4
12
=
1
3
6. 3, 3 , 3 , ...
2 4
a = 3, r =
3
=
1
2
a = 2, r =1
2
Sn
= a(r n 1)
r 1
S6
=
2[( )
6 1]
1
=
63
16
= 3
15
16
12
12
32
Sn
= a(r n 1)
r 1
S6
=
3[( )
6 1]
1
=
63
32
= 1
31
32
12
12
Sn
=a(r n 1)
r 1
S6
=
12[( )
6 1]
1
=
1456
81
= 17
79
81
13
13
2, 6, 18, ...
a = 3, r =6
3
= 2
Sn
= a(r n 1)
r 1
S6
=3(26 1)
2 1
= 189

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E Find the sum of all the terms for each of the geometric progressions below.
5, 10, 20, ..., 640
3. 2, 6, 18, ..., 1 4582. 2, 8, 32, ..., 2 048
Tn
= 2048
2(4)n 1 = 2048
4n 1 = 1024
4n 1 = 45
n 1 = 5
n = 6
Sn
= a(r n 1)
r 1
S6
=2(46 1)
4 1
S6 = 2 730
Tn
= 1458
2(3)n 1 = 1458
3n 1 = 729
3n 1 = 36
n 1 = 6
n = 7
5. 1,
1,
1, ...,
1
2 4 128
4. 2.4, 4.8, 9.6, ..., 38.4
Example
Tn
= 640
5(2)n 1 = 640
2n 1 = 128
2n 1 = 27
n 1 = 7
n = 8
Sn
=a(r n 1)
r 1
S8
=5(28 1)
2 1
S8
= 1 275
Tn
= 2 916
4(3)n 1 = 2 916
3n 1 = 729
3n 1 = 36
n 1 = 6
n = 7
Sn
=a(r n 1)
r 1
S7
=4(37 1)
3 1
S7
= 4 372
Sn =a(rn
1)
r 1
S5
=2.4(25 1)
2 1
S5
= 74.4
Tn = 38.42.4(2)n 1 = 38.4
2n 1 = 16
2n 1 = 24
n 1 = 4
n = 5
Sn
=a(r n 1)
r 1
S7
=2(37 1)
3 1
S7
= 2 186
12
Tn
=1
128
( )n 1
=1
128
( )n 1 = ( )
7
n 1 = 7
n = 8
12
12
Sn
= a(r n 1)
r 1
S8
=
1[( )
8
1 ]
1
= 1
127
128
121
2
1. 4, 12, 36, ..., 2 916
r =12
4
= 3r =
10
5
= 2
r =8
2
= 4
r =6
2
= 3
r =4.8
2.4
= 2
12
r =
=
1
1
2

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T3
= 60; T5
= 240
ar 2 = 60 1
ar 4 = 240 2
2: ar 4
=240
1 ar 2 60
r 2 = 4
r = 2
a(24) = 240
a = 15
S6
= a(rn 1)
r 1
=
15(26 1)
2 1
= 945
2. The sum of the first three terms of a geometric
progression with a common ratio of1
is 780.
3Find
(a) the first term,
(b) the sum of all the terms from the third term to
the eighth term.
13
13
13
13
S8
S2
=
540( 1 ( )8
)
540(1 ( )2
)1 1
= 80971
720
81
= 8971
81
3. In the progression 1, 3, 9, 27, ..., find the least
number of terms required such that its sum exceeds
1 000.
r = 3; a = 1
Sn 1 000
1(3n 1) 1 000
2
3n 1 2 000
3n 2 001
nlog 3 log 2 001n 6.919
n = 7
1. In a geometri progression, the third term and the
fifth term are 60 and 240 respectively. Find the
sum of thefi
rst six terms.
F Solve the following problems.
S3
= 780; r =1
3
a[1( )
3
]= 780
1 1
3
a
(26
)
= 520 27 a = 540
13
(a)
(b)
4. The first term and the common ratio of a geometric
progression are 27 and4
respectively. Find the
3
smallest value ofn such that the sum to the n terms
exceeds 891.
a = 27, r =4
3
43
43
43
43
27[ ( )n
1] 891
1
( )n
1 11
( )n
12
nlog4 log 12
3
n 8.64
n = 9
The least number of terms required is 7.
The smallest value ofn is 9.

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13
G Estimate the sum to infinity of the following geometric progression.
1. 20, 10, 5, ...Example
243, 81, 27, 9, ...
a = 243, r =81
=1
243 3
S =243
1 1
3
= 243
3
2
= 364.5
Tip
S = a
1 r
where r = 0
a = 20; r =1
2
S =20
1 1
2
= 20 2
= 40
3. 4, 0.4, 0.04, 0.004, ...
a = 4; r = 0.1
S =4
1 0.1
=
4
0.9
= 4
4
9
2. 27, 18, 12, ...
a = 27; r =2
3
S =27
1 2
3
= 27 3
= 81
5. Estimate the sum to infinity of the following series
9 + 3 + 1 +1
+1
+ ...
3 9
4. 16, 10
2, 7
1, ...
3 9
a = 16; r =
S =16
1 2
3
= 48
102
3
16=
2
3
a = 9; r =3
=1
9 3
S =9
1 1
3
= 13
1
2
6. A geometric progression with 28 as its first termhas the sum to infinity of 70. Find its common
ratio.
7. The first term of a geometric progression is 12 andthe sum to infinity is 24. Find its common ratio.
a = 28; S = 70
S =a
1 r
70 =28
1 r
70 70r = 28
70r = 42
r =
42=
3
70 5
a = 12; S = 24
24 =12
1 r
24 24r = 12
24r = 12
r =
12

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H Write the following recurring decimal numbers as a single fraction in its lowest terms.
Example
0.333... = 0.3 + 0.03 + 0.003 + ... 1. 0.6666...
0.6666 = 0.6 + 0.06 + 0.006 ...
a = 0.6, r =0.06
0.6
= 0.1
S =0.6
1 0.1
=
0.6
0.9
= 2
3
3. 0.459459459 ...2. 0.131313 ...
0.131313 = 0.13 + 0.0013 + 0.000013 ...
a = 0.13, r =0.0013
0.13
= 0.01
S =0.13
1 0.01
=
0.13
0.99
=
13
99
a = 0.3, r =0.03
0.3
= 0.1
S = a
1 r
=
0.3
1 0.1
=
0.3
0.9
=
3
9
=
1
3
0.459459459 = 0.459 + 0.000459 ...
a = 0.459, r =0.000459
0.459
= 0.001
S =0.459
1 0.001
=
0.459
0.999
=
17
37
5. 0.151515 ...4. 0.555 ...
0.555... = 0.5 + 0.05 + 0.005 + ...
a = 0.5, r =0.05
0.5
= 0.1
S =0.5
1 0.1
=
0.5
0.9
=
5
9
0.151515 = 0.15 + 0.0015 + 0.000015
a = 0.15, r = 0.0015
0.15
= 0.01
S =0.15
1 0.01
=
0.15
0.99
=
15
99
=
5
33

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SPM Practice1
1. The sequence 10, 7, 4, ..., 134, 137 is an
arithmetic progression. Find
(a) the number of terms in the progression,
(b) the sum of all the terms in the progression.
2. The sum of the first n terms of an arithmetic
progression is given by Sn
= 6n n2. Find
(a) the fourth term,
(b) the sum of all the terms from the fifth term to
the tenth term.
3. Given that the first term and the fifth term of anarithmetic progression are 6 and 10 respectively.
Find
(a) the common difference,
(b) the tenth term,
(c) the sum of the first ten terms.
4. The sum of the first eight terms of an arithmetic
progression is 132. The fourth term of the
progression is 17. Find
(a) the common difference and the second term,
(b) the sum of all the terms from the 9th term to
the 25th term.
5.
The diagram shows several circles where the radii
of the circles decrease by 1 unit consecutively.
Show that the circumference of the given circles
form an arithmetic progression. Hence, find the
total circumference of the first 10 circles.
6. Given that the first and the fourth terms of a
geometric progression are 12 and32
respectively.
9
Find
(a) the common ratio,
(b) the sum of the first 8 terms.
7. Given that the second and the fourth terms of a
geometric progression are 3 and 27 respectively,
and the common ratio is positive. Find
(a) the common ratio,
(b) the first term,
(c) the seventh term.
8. The amount of insurance coverage for a new car
in the year 2006 was RM80 000. The sum insured
for each following year will decrease by 8% as
compared to the previous year. Calculate the sum
insured for the year 2014.
9. The second term of a geometric progression
exceeds the first term by 8 while the fifth term
exceeds the fourth term by 216. Find the sum of
the first ten terms.
10.
Asraf stacked up 50sen coins as shown in the
diagram. The sequence of the number of coins
in each stack forms a geometric progression.
Calculate
(a) the number of coins in the 10th stack,(b) the sum of money, in ringgit Malaysia, for the
first ten stacks of coins.
11. P = x + 3 +3
+3
+ ... is a series formed from a
4 16
geometric progression with an infinite number of
terms. Find
(a) the value ofx,
(b) the value ofP.
12. The sum of the first 5 terms of an arithmetic
progression is 30 and the sum of the next 5 terms
is 80. Find
(a) the first term and the common difference,
(b) the sum of all the terms from the 11th term to
the 20th term.
r r 1r 2
1 coin 3 coins 9 coins
50c50c
50c