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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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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 50-sen 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