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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