PROGRAM DIDIK CEMERLANG AKADEMIK

SPM

PROGRESSIONS (Geometric Progression)

ORGANISED BY:

JABATAN PELAJARAN NEGERI PULAU PINANG

ADDITIONAL MATHEMATICSFORM 5

MODULE 2

CHAPTER 2 : GEOMETRIC PROGRESSIONS

Contents Page

2.1 CONCEPT MAP (GEOMETRIC PROGRESSIONS)2

2.2 IDENTIFY CHARACTERISTICS OF GEOMETRIC PROGRESSIONS

2.3 DETERMINE WHETHER A GIVEN SEQUENCE IS A GEOMETRIC PROGRESSION

2.4 DETERMINE BY USING FORMULA

a) specific terms in geometric progressions b) The number of terms in geometric progressions

2.5 FIND

a) The sum of the first n terms of of geometric Progressions b) The sum of a specific number of consecutive terms of geometric Progressions c) The value of n ,given the sum of the first n terms of geometric Progressions

2.6 FIND a) The sum to infinty of geometric progressions

b) The first term or common ratio, given the sum to infinity

2.7 SOLVE PROBLEMS INVOLING GEOMETRIC PROGRESSIONS

3

4

6

6

11

14

SPM Questions 15

Assessment test18

Answers 20

2

CHAPTER 2 : GEOMETRIC PROGRESSION

2.1 CONCEPT MAP

3

PROGRESSION

ARITHMETICPROGRESSION

GEOMETRIC PROGRESSION

Tn = arn-1

.a = first term

.r = Common ratio

.n = number of terms

SUM OF THE FIRST n TERMS

Sn = 1

)1(

−−

r

ra n

, r > 1

OR

Sn = r

ra n

−−

1

)1(, r < 1

THE n th TERM Tn

THE r th TERM Tr

T r = S r - S 1−r

SUM OF INFINITY

Sn = r

a

−1, -1 < r < 1

PROGRESSIONS( Geometric Progression)

a = _______________r = _______________l = _______________n = _______________

2.2 Identify characteristics of geometric progression:

EXERCISE 1:Complete each of the sequence below when give a (fist term ) and r (common ratio) .

A .r the first four terms of geometric progression,

Example:a) -3 2 -3, (-3)(2)1 = - 6, (-3)(2)2 = -12, (-3)(2)3 = -24

.b) 3 -2

.c) 4 3

.d) -6 -2

.e) 2

1

3

1

.f) 3

1y

3

4y

4

GEOMETRIC PROGRESSION

n

n

The :

T = ___________

or

T = ___________

thn termn

n

The sum of the first n term:

S = ______________

or

S = ______________

Fill in the blank

2.3 Determine whether a given sequence is an geometric progression

EXERCISE 2 :

2.3.1 Determine whether a given sequence below is an geometric progression.

Example : a) -8, 4, -2,……

.r = common ratio

.r = 8

4

−=

2

1

− ,

4

2−=

2

1

− (true)

b) 5, 11, 17, 23,…

c) 16, -8, 4,…..

d) -20, -50, -30, -35,…..

e) 3

1x,

9

2x,

27

4x……

f) a5, a4 b, a3 b2

g) 4

1,

12

1,

36

1,….

h) 27

1,

9

1,

3

1,…

5

EXERCISE 3 :

2.3.2 Given that the first three terms of a geometric progression are below. Find the value of x

Geometric progression Value xExample : a) .x, x + 4, 2x + 2,… x

x 4+=

4

22

++

x

x

(x + 4)2 = x(2x + 2).x2 + 8x + 16 = 2x2 + 2x

0 = x2 – 6x – 190 = (x – 8)(x + 2).x – 8 = 0 @ x + 2 = 0

Hence x = 8 @ x = -2

b) x, x + 2 , x + 3

c) x + 3, 5x - 3, 7x + 3

e) x – 6, x, 2x + 16

6

2.4 Determine by using formula:

EXERCISE 4:

2.4.1 specific terms in arithmetic progressions

Example :1. Find the 7 th term of the geometric

progression. - 8, 4 , -2 , …..Solution:

a = - 8 r = 8

4

−=

2

1

−

T7 = (-8)(2

1

−)7-1

= 8

1

−

2. Find the 8 th term of the geometric progression.

16, -8, 4,…

3. For the geometric progression

9

4,

3

2, 1 , ….. ,find the 9 th term .

4. Find the 3 th term of the geometric progression 50, 40, 32…….

5. Find the 10 th term of geometric progression a5, a4 b, a3 b2…………

6. Given that geometric progression 5.6, 1.4, 0.35, …… Find the 10th term.

7

Tn = arn-1

2.4.2 Find the number of terms of the arithmetic progression

EXERCISE 5:

Example :

a) 64, - 32, 16,…….-8

1

Tn = -8

1 a = 64, r = -

2

1

64

−

2

11−n = -

8

1

−

2

11−n =

64

1

−

8

1

−

2

11−n =

−

2

16

−

2

13

−

2

11−n =

−

2

19

.n - 1 = 9 .n = 10

b) 2, -4, 8, ……512

. c) 405, - 135, 45, …… - 27

5.d)

9

4,

3

2, 1……..

16

81,

8

2.5 Finda) The sum of the first n terms of geometric progressionsb) The sum of a specific number of consecutive terms of

geometric progressionsc) The value of n , given the sum of the first n terms of

geometric progressions.

2.5.1 Find the sum of the first n terms of geometric progressions

EXERCISE 6:

geometric progressions Find the sum of the first n term

Example :

.a) 2, - 4, 8,…………

.a = 2 r = 2

4− = - 2

S7 = )2(1

))2(1(2 7

−−−− = 86

b) 5, 10, 20……….. S 8

c) 12, -6, 3……… S 9

d) 3

1x,

9

2x,

27

4x,….

S 6

9

Sn =

1

)1(

−−

r

ra n

, r > 1

OR

Sn = r

ra n

−−

1

)1( , r < 1

2.5.2 The sum of a specific number of consecutive terms of geometric progressions

EXERCISE 7:

geometric progressions Find the sum of the first n term

Example 1

.a) 4, 2, 1,………… 64

1

.a = 4 r = 4

2 =

2

1

Tn = 64

1

4

2

11−n

= 64

1

2

11−n =

4

1

64

1

2

11−n =

2

12

2

16

.n – 1 = 8 .n = 9

S 9 = ( )[ ]( )2/11

2/114 9

−−

= 764

63

3 1, 3, 9,……. 2187.

c) 24, 12, 6, …….4

3

10

2.5.3 The value of n , given the sum of the first n terms of geometric progressions.

EXERCISE 8 :

Example :

.a) The first and 4th tems of a geometric progression are 1

2 and

27

128.

Find the value of rSolution :

T 4 = 27

128

1

2 (r 3 ) =

27

128

(r 3 ) = 27

128

2

1

(r 3 ) = 27

64 , Hence r =

3

4

b) The first and 6th tems of a geometric progression are 21

2 and 607

1

2.

Find the value of r

c) The common ratio and 5th tems of a geometric progression are 2

3 and 7

22

27.

Find the value of a

11

2.4 Find :a) the sum to infinity of geometric progressions

b) the first term or common ratio, given the sum to infinity of geometric progressions.

EXERCISE 9:

Find the sum to infinity of a given geomertric progression below:

Example:

2 26, 2, , ,.......

3 9− −

a = 6

2 1

6 3r

−= = −

16

= 1

1- -3

9 =

2

aS

r∞ =−

1. 24, 3.6, 0.54, …….

2. 81, -27,9, ……..

3. 1 1 1

, , ,.......2 4 8

..

EXERCISE 10:

12

1

aS

r∞ =−

sum to infinity

a = first term

r = common ratio

S∞ =

1. The sum to infinity of a geometric progression is 200. Given that the first term is 52. Find the common ratio.

2. Given that the common ratio of

geometric progression is1

25. The

sum of the first n terms,where n is large enough such that 0nr = is 75. Findthe first term.

Example:The sum to infinity of a geometric

progression is 8. Given that the first term is 2. Find

a) the common ratiob) the third term

Solution: a)

8

81

28

13

r= 4

S

a

r

r

∞ =

=−

=−

b)

1

3 13

=24

9 =

8

nnT ar −

−

=

3. The sum to infinity of a geometricprogression is 600 and the common ratio is 0.4 . Find

a) the first term b) the minimum number of

terms such that the sum of terms to be more then 599.

13

4. Express each of the recurring decimal below as a fraction in its simplest form.

Example:

0.3……

0.3…= 0.3 + 0.03 + 0.003 + …..

0.3

1 0.10.3

0.91

3

=−

=

=

a) 0.444…….

b) 0.232323……

Example:

4.020202……

4.020202... 4 0.02 0.0002 0.000002 ...

0.02 = 4 +

1-0.010.02

= 4 + 0.99

2 =4

99

= + + + +

c) 1.121212…..

d) 5.070707…...

14

2.5 Solve problems involving geometric progressions:

EXERCISE 11:

Example: A garderner has a task of cutting the grass of a lawn with an area of 1000 2m .

On the first day, he cut an area of 216m . On each successive day, he cuts an area

1.1 times the area that he cut the previousday the task is completed. Find

a) the area that is cut on the 10th day.b) The number of the days needed to

complete the task.

Solution:a) a=16 , r=1.1

1

10

2

16(1.1)9

= 37.73 m

nnT ar

T

−==

n

n

n

n

b) 1000

16 1.1 -1 1000

1.1-1

160(1.1 1) 1000

1.1 1 6.25

1.1 > 7.25

log 1.1 log 7.25

log1.1 log 7.25

log7.25 n >

log1.1

n >20.79

The smallest integer valu

n

n

S

n

e of n is 21

hence,the number of the days needed to complete

the task is 21 days.

Osman is allowed to spend an allocation of RM1 million where

the maximum withdrawal each day must not exceed twice the amount withdrawn the day before. If Osman withdraws RM200 on the first day, determine after how many days the amount of money allocated will all be used up.

15

SPM QUESTIONS:

1. 2003 (Paper 1: No.8)

In a geometric progression, the first term is 64 and the fourth term is 27. Caculate(a) the common ratio(b) the sum to infinity of the geometric progression. [4 marks]

2. 2004(Paper 1: No.9)

Given a geometric progression 4

,2, , ,.....y py

,express p in terms of y.

3. 2004(Paper 1: No.12 )

Express the recurring decimal 0.969696……as a fraction in its simplest form. [4 marks]

16

4. 2004(Paper2: Section A: No.6)

Diagram 2 shows the arrangement of the first three of an infinite series of similartriangles. The first triangle has a base of x cm and a height of y cm. The measurements of the base and height of each subsequent triangle are half ofthe measurements of its previous one.

Diagram 2

(a) Show that the areas of the triangles form a geometric progression and statethe common ratio. [3 marks]

(b) Given that x= 80 cm and y= 40 cm,

(i) determine which triangle has an area of 216 cm

4,

(ii) find the sum to infinity of the areas, in 2cm , of the triangles. [5 marks]

5. 2005 (Paper 1 : No.10)

17

y cm

xcm

The first three terms of a sequence are 2 , x , 8Find the positive value of x so that the sequence is(a) an arithmetic progression(b) a geometric progression [4 marks]

6. 2005 (Paper 1: No. 12)

The sum of the first n terms of the geometric progression 8,24,72,….is 8744.Find(a) the common ratio of the progression(b) the value of n [4 marks]

ASSESSMENT:

18

1. The first three terms of a geometric progression are 2x + 3, x and x – 2 with a common ratio r , where -1 < r < 1. Find

(a) the value of x (b) the sum of the first n terms ,where n is large enough such

that 0nr

2. In the progression 5 , 10 , 20 , 40 , ……. Find the least number ofterms required such that their sum exceeds 1000.

3. The third term and the sixth term of a geometric progression are27 and 8 respectively. Find the second term.

4. In a geometric progression, the sum of the first five terms is 31

8 .

19

Given that the common ratio is 1

2 . Find

(a) the first term(b) the sum of all the terms from the fourth to the sixth term.

5. The third term of a geometric progression exceeds the second termby 6 while the fourth term exceeds the third term by 2. Find thesum of the first 5 terms.

ANSWERS:EXERCISE 1: EXERCISE 2:

20

b) 3, 6, -12, 24c) 4, 12, 36, 108d) -6, 12, 124, 48

e)1

2,

1

6,

1

18,

1

54

f)3

y,

24

9

y,

316

27

y,

464

81

y

a) trueb) falsec) trued) falsee) truef) trueg) falseh) true

EXERCISE 3:b) x = -4c) x = 3d) x = -12 @ x = 8

EXERCISE 4:

1. T 7 = - 1

8

2. T 8 = - 1

2

3. T 9 = 26344

23044. T 3 = 32

5. T10 = 9

4

b

a6. T10 = 0.000021

EXERCISE 5:b) n = 9c) n = 8d) n = 7

EXERCISE 6:b) 275

c) 81

64

d)665

729EXERCISE 7:

c) n = 8, S 8 = 3280d) n = 9, S 9 = -1022

e) n = 6, S 6 = 471

4

EXERCISE 8:b) r = 3c) a = 3

EXERCISE 9: 1. 28.24

2. 3

604

3. 1

EXERCISE 10:2. r=0.74 3. a=724. a) 360 b) 7

5.

4a)

923

) 99

4) 1

337

) 599

b

c

d

21

EXERCISE 11: 13 days

SPM QUESTION:

1. a) 3

4r b) 256S

2. 2

8p

y

3.32

33

4. a) 1

4r

b)i. n=5 ii. 1

21333

5. a) x=5 b) x=4 6. a) r=3 b) n=7

ASSESSMENT:

1. a) x=3 b) 27

22. 8

3. 1

402

4. a) -2 b) 7

16

5. 1

403

22