geometric progressions.docx

7/29/2019 Geometric Progressions.docx

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CHAPTER 2 : GEOMETRIC PROGRESSION

2.1 CONCEPT MAP

PROGRESSION

ARITHMETICPROGRESSION

GEOMETRICPROGRESSION

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 rth TERM Tr

Tr

= Sr- S 1r

SUM OF INFINITY

Sn =

r

a

1, -1 < r < 1


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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)21

31

.f)3

1y

3

4y

GEOMETRIC PROGRESSION

n

n

The :

T = ___________

or

T = ___________

thn term

n

n

The sum of the first n term:

S = ______________

or

S = ______________

Fill in the blank


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2.3 Determine whether a given sequence is an geometricprogression

EXERCISE 2 :

2.3.1 Determine whether a given sequence below is an geometricprogression.

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,


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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 x

Example :

a) .x, x + 4, 2x + 2,x

x 4=

4

22

x

x

(x + 4)2 = x(2x + 2)

.x2 + 8x + 16 = 2x2 + 2x0 = 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


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2.4 Determine by using formula:

EXERCISE 4:

2.4.1 specific terms in arithmetic progressions

Example :

1. Find the 7th

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 8th

term of the geometric

progression.

16, -8, 4,

3. For the geometric progression

9

4,

3

2, 1 , .. ,find the 9

thterm .

4. Find the 3th

term of the geometricprogression

50, 40, 32.

5. Find the 10th

term of geometric

progression a5, a4 b, a3 b2

6. Given that geometric progression5.6, 1.4, 0.35,

Find the 10th

term.

Tn = arn-1


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2.4.2 Find the number of terms of the arithmetic progressionEXERCISE 5:

Example :

a) 64, - 32, 16,.-8

1

Tn = -8

1a = 64, r = -

2

1

64

2

1 1n= -

8

1

2

1 1n=

64

1

8

1

2

1 1n=

2

1 6

2

1 3

2

1 1n=

2

1 9

.n - 1 = 9.n = 10

b) 2, -4, 8, 512

. c) 405, - 135, 45, -27

5.d)

9

4,

3

2, 1..

16

81,


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2.5 Finda)The sum of the first n terms of geometric progressionsb)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 ofgeometric progressions.

2.5.1 Find the sum of the first n terms of geometric progressionsEXERCISE 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.. S8

c) 12, -6, 3 S9

d)3

1x,

9

2x,

27

4x,.

S6

Sn =1

)1(

r

ra n, r >1

OR

Sn =r

ra n

1

)1(, r


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2.5.2 The sum of a specific number of consecutive terms of geometricprogressions

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

1 1n=

64

1

2

1 1n=

4

1

64

1

2

1 1n=

2

1 2

2

1 6

.n 1 = 8.n = 9

S9

= 2/11

2/1149

= 7

64

63

3 1, 3, 9,. 2187.

c) 24, 12, 6, .4

3


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

EXERCISE 8 :

Example :

.a) The first and 4th tems of a geometric progression are1

2and

27

128.

Find the value of rSolution :

T 4 =27

128

1

2(r

3) =

27

128

(r3) =

27

128

2

1

(r3) =

27

64, Hence r =

3

4

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

2and 607

1

2.

Find the value of r

c) The common ratio and 5th tems of a geometric progression are2

3and 7

22

27.

Find the value of a


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2.4 Find :a) the sum to infinity of geometric progressionsb) the first term or common ratio, given the sum to

infinity of geometric progressions.

EXERCISE 9:

Find the sum to infinity of a givengeomertric progression below:

Example:

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

3 9

a = 6

2 1

6 3r

1

6=

11- -

3

9=

2

aS

r

1.

24, 3.6, 0.54, .

2. 81, -27,9, ..

3.1 1 1

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

..

1

aS

r

sum to infinity

a = first term

r = common ratio

S


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EXERCISE 10:

1. The sum to infinity of a geometricprogression is 200. Given that the

first

term is 52. Find the common ratio.

2. Given that the common ratio ofgeometric progression is

1

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. Finda) the common ratiob) the third term

Solution:a)

8

81

28

1

3r=

4

S

a

r

r

b)1

3 1

3=24

9=

8

n

nT ar

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

ratio is 0.4 . Find

a) the first termb) the minimum number of

terms such that the sum of

terms to be more then 599.


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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 0.1

0.3

0.9

1

3

a) 0.444.

b) 0.232323

Example:

4.020202

4.020202... 4 0.02 0.0002 0.000002 ...

0.02= 4 +

1-0.01

0.02= 4 +

0.99

2=4

99

c) 1.121212..

d) 5.070707...


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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 10002m .

On the first day, he cut an area of216m .

On each successive day, he cuts an area1.1times 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

n

nT arT

n

n

n

n

b) 1000

16 1.1 -11000

1.1-1

160(1.1 1) 1000

1.1 1 6.251.1 > 7.25

log 1.1 log 7.25

log1.1 log7.25

log7.25n >

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 wherethe maximum withdrawal each

day must not exceed twice theamount withdrawn the day

before. If Osman withdrawsRM200 on the first day,

determine after how many days

the amount of money allocatedwill all be used up.


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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 p

y,express p in terms of y.

3. 2004(Paper 1: No.12 )Express the recurring decimal 0.969696as a fraction in its simplest form.

[4 marks]


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4. 2004(Paper2: Section A: No.6)Diagram 2 shows the arrangement of the first three of an infinite series of similar

triangles. The first triangle has a base of x cm and a height of y cm. Themeasurements of the base and height of each subsequent triangle are half of

the measurements of its previous one.

Diagram 2

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

(b) Given that x= 80 cm and y= 40 cm,(i) determine which triangle has an area of

216 cm4

,

(ii) find the sum to infinity of the areas, in2cm , of the triangles. [5 marks]

y cm

xcm


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5. 2005 (Paper 1 : No.10)The first three terms of a sequence are 2 , x , 8

Find 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]


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ASSESSMENT:

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.


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4. In a geometric progression, the sum of the first five terms is 318

.

Given that the common ratio is1

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 the

sum of the first 5 terms.


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ANSWERS:

EXERCISE 1:

b) 3, 6, -12, 24

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

e) 12

,1

6,

1

18,

1

54

f)3

y,

24

9

y,

316

27

y,

464

81

y

EXERCISE 2:

a) trueb)

falsec) true

d) falsee) truef) trueg) falseh) true

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

EXERCISE 4:

1. T 7 = - 18

2. T 8 = -1

2

3. T9

=26344

2304

4. T3

= 32

5. T 10 = 94ba 6. T 10 = 0.000021

EXERCISE 5:

b)n = 9c) n = 8d)n = 7

EXERCISE 6:

b)275c) 8 1

64

d) 665729

EXERCISE 7:c) n = 8, S

8= 3280

d)n = 9, S9

= -1022

e) n = 6, S6

= 471

4

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

EXERCISE 9:1. 28.24

2.3

604

3. 1

EXERCISE 10:2. r=0.743. a=724. a) 360 b) 7


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5.

4a)

9

23)

99

4) 1

337

) 599

b

c

d

EXERCISE 11:13 days

SPM QUESTION:

1. a) 34

r b) 256S

2. 28p y 3. 32

33

4. a) 14

r

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

2

2. 8

3.1

402

4. a) -2 b)7

16

5.1

403

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