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TTooppicc

66LEARNING

Sequencesand Series

OUTCOMES

By the end of this topic, you should be able to:1. Describe the basic concepts of series, sequences, as well as convergent

and divergent series;

2. Explain how to solve problems involving arithmetic and geometricprogressions or series;

3. Illustrate the Binomial Theorem; such as to expand (a+b)nwherenis a positive integer and to expand (1+x)n, wherenis a rational

number and|x|

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100 TOPIC6 SEQUENCESANDSERIES

Lastly, you will learn about The Binomial Theorem. Some examples and exercisesare given throughout this topic for you to practise the concepts covered.

Before we go further, can you give at least three reasons why students shouldlearn about this topic? How can this knowledge be applied in a real-life situation?You may look this up in the Internet and share it with your peers in myVLE.

6.1 INTRODUCTIONTOSEQUENCESANDSERIES

Do you know what a sequence is? A sequence is a set of listed numbers. Thesenumbers are known as terms and they have a common characteristic. The firstnumber is known as the first term, the second number as the second term, and so

on.

Let us consider the following sequences;

1, 3, 5, 7, 9, 11, ........ is a sequence of positive odd numbers.

2, 4, 6, 8, 10, 12, ...... is a sequence of a positive even numbers.

0, 1, 2, 3, 4, 5, 6, .......is a sequence of a non-negative integers.

The terms in a sequence are normally represented using alphabets such asa1,c

1

orT1, which is the first term.a

2,c2,T2represents the second term and so on.

Thus the sequences above can be represented as follows:

a1,a2,a3,........

nor

c1,c

2,c

3, ...,c

nor

T1,T

2,T

3, ...,T

n

These are the three common variables used torepresent the terms in a sequence. In fact, any othervariables can also be used.

For example, if the sequence consists of positive odd numbers, the first term isa1=1, the second term,a2=3, the third term,a3=5 and so on.

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TOPIC6 SEQUENCESANDSERIES !101

The sum of the numbers in a sequence is known as aseries. Therefore, from theprevious sequences, we will have the series as follows:

1+3+5+7+9+11+......... is a series of a positive odd numbers.

2+4+6+8+10+12+...... is a series of a positive even numbers.

0+1+2+3+4+5+6+..... is a series of a positive integers.

From the first series above we can consider the sum of the firstnterms:

The sum of the first three terms=a1+a2+a3=1+3+5=9.

or,

The sum of the first five terms = a1+a

2+a

3+a

4+a

5

=1+3+5+7+9=25

A sigma notation is used to represent the summation of numbers in a series.A variable,Swith a subscripti, is used to represent the sum of the firstnterms.

xnis the term

nis the term number

Thus, the summation above can also be written as follows:

The sum of the first three terms,S3=!ai=i=1

=

a1+a2+a3

1+3+5

or,

The sum of the first five terms,S5=!

aii

=1

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=9 =a1+a

2+a

3+a

4+a

5

=1+3+5+7+9

=25

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Example 6.1:Given the series; 1+2+3+4+5+6+......+n, find:

(a) The fourth term

(b) The sum of the first six terms

Solution:

(a) The fourth term,a4=4.

(b) The sum of the first six terms

S6=

!an=

n=1

=

a1+a2+a3+a4+a5+a6

1+2+3+4+5+6=21

Example 6.2:

Find the sum!2n1#n=0

Solution:Sinceibegins from 0 to 3, we have the summation of the first four terms instead

of just three terms.

!2n1#=n=0

=

[2(0) 1]+[2(1) 1]+[2(2) 1]+[2(3) 1]

1+1+3+5=8

Example 6.3:

Find the first four terms and the sum for the sequencec1,c

2,c

3,........ if

cn=3n+7.

Solution:Write down the first four terms as follows:

The first term,

The second term,

c1=3(1)+7=3+7=10

c2=3(2)+7=6+7=13

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The third term,

The fourth term,

c3=3(3)+7=9+7=16

c4=3(4)+7=12+7=19

Thus, the first four terms of the sequence are 10, 13, 16, 19.

The sum of the first four terms=!

cii

=1

=c1+c

2+c

3+c

4

=10+13+16+19

=58

Example 6.4:Write down the first five terms of the sequence defined as follows:

a1=2,a

2=3,a

n=a

n-1+a

n-2+1

Solution:The first two terms are already given. We can use the general formula given tofind the next three terms as follows:

Whenn=3;

Whenn=

4;

Whenn=

5;

a3=a2+a1=3+2=5

a4=a

3+a

2=5+3=8

a5=a

4+a

3=8+5=13.

The first five terms of the sequence are 2, 3, 5, 8, 13.

6.1.1 UseofExplicitoRec!si"e#o$!l%to#i&'S!ccessi"eTe$sfo%Se(!e&ce

For any given sequence, we can try to find a general representation of its terms.

We can solve such problems by constructing a table and trying to observe anypatterns, or rules on how those terms in the sequence are obtained.

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Let us consider the following examples.

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10) TOPIC6 SEQUENCESANDSERIES

Example 6.5:Write in notation form, the general representation of the sum of the firstntermsin the sequence:

1, 3, 5, 7, 9, 11,

Solution:

i aiTerm Polar

1 1 2%1 1

2 3 2%2 1

3 5 2%3 1

4 7 2%4 1

5 9 2%5 1

n an

2%n1

We construct the table above. The value ofiis written in the first column. Theterms of the sequence are written accordingly in the second column. The thirdcolumn in the table is very important. We try to observe a specific rule orpolar/pattern on how these terms are obtained.

Thus, the general representation, i.e. thenthterm of the sequence is 2n1.

Hence, the sum of the firstnterms of the sequence is! 2i1#.i=1

Example 6.6:Write in notation form, the general representation of the sum of the firstntermsin the sequence:

1, 4, 9, 16, 25,

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TOPIC6 SEQUENCESANDSERIES !10*

Solution:By constructing the following table, we try to observe any suitable pattern howterms are obtained.

i aiTerm Polar

1 1 12

2 4 22

3 9 32

4 16 42

5 25 52

.. .

n an n2

Thus, the general representation, i.e. thenthterm of the sequence isn

2.

Hence, the sum of the firstnterms of the sequence is!i2.i=1

6.1.2 Co&"e+e&ceoDi"e+e&ceof%Seies

In mathematics, when we talk aboutconvergence, it is defined as the property ormanner ofapproaching a limit such as a point, a line, a function or a value.Whereas,divergence is defined as the property or manner whichfails toapproach a limit.

(Source: The free dictionary.com)

Before we discuss the convergence or divergence of any given series, let us

consider an infinite series. An infinite series is written as below:

S=u1+u

2+u

3+........+u

n+...........

i.e. the sum of thenterms of an infinite sequence.

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To consider the convergence of such a series, i.e. whether it adds up to a finitequantity, we will introduce some definitions. Firstly, let us assume that the series

is finite (approaching a limit) and define thenthpartial sum of the series as the

sum of the firstnterms:

Sn=sum ofnterms=u

1+u

2+u

3+........+u

n

Then the infinite series can be regarded as thisnthpartial sum asntends to

infinity, i.e.

S=sum toinfinity=limSn

IfS=limS=l, a definite value, then we say that the series converges to then&

value l. Otherwise, the series is divergent.

Using the sigma notation, we write:

Sn=!urr=

1

S=!urr=

1

Clearly, whether or not a series converges depends on the termsurand also on

their relation to each other, i.e. inur1. This last quantity appears in the ratio

test for convergence.

One way of testing for convergence, of course, is to investigate the limit of thenth

partial sum directly.

Example 6.7:Investigate the convergence of the geometric series

S=a+ar+ar2+ar

3+........+ar

n-1+.........

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TOPIC6 SEQUENCESANDSERIES !10,

Solution:We have:

Sn=a+ar+ar2

+........+arn-1

Sn=a1$r n #

=1$r ( a) n

,1r

Now, 1

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And therefore the arithmetic series is divergent.

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10- TOPIC6 SEQUENCESANDSERIES

You have now looked at how to define whether a series is convergent ordivergent. Besides the examples given above, you may also demonstrate thisknowledge with your students using this method shared by Khan Academy.

View the video (or encourage your students to view the video) here:https://www.khanacademy.org/math/integral-calculus/sequences_series_ approx_calc/seq-conv-diverg/v/convergent-and-divergent-sequences

6.1.3 TestsfoCo&"e+e&ce

As noted in the previous section, if we can write the general form forSn, then we

can find limS and check the convergence directly, but this is not alwaysn&

possible. Fortunately, a number of different tests have been developed todetermine whether a series converges by considering its general form.

Before we even start to discuss the convergence of a series, we must first checkthatlimu10, because if this is not so then the series cannot converge.

n&

Assuming the series converges, then from:

un1S

n S

n1

we have;limSS # 101limu

n& n&

So, if the series converges,limu10.n&

However, the converse is not true, i.e.limu10 , which does not imply that then&

series converges.

Having verified thatlimu10, we can continue to apply the tests forn&

convergence given as follows.

First, write the general form forSn.

(a) In studying the convergence of a series, it is only theinfinite tailwhich isimportant not an initial infinite number of terms.

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(b) Some of the tests given below apply only to series whose terms alleventually become positive.

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TOPIC6 SEQUENCESANDSERIES !10.

The various types of tests for convergence are stated below:

(a) The Comparison Test

This is essentially a generalisation of the method we used for the harmonicseries as follows:

Thenthpartial sum of the harmonic series:

S 1 1 21

2

2 1

3

2 1

4

2 .......... 2 1

n

2 ........

Is

Sn 1 1 2

1

2

2 1

3

2 .......... 2 1

n

Suppose we wish to study the convergence of the series of positive terms:

U 1 u1

2 u2

2 .........2 un 2 ........

Then in the comparison test we compare this with a known series. Thus, let

V 1 v1

2 v2

2 .........2 vn 2 ..........

be a known standard series. Then ifur3cv

r,c=aconstant andV

is convergent, then so isU. Ifur.cv

randVdiverges, then so doesU.

This test is intuitively obvious.

Example 6.9:Apply the comparison test to the series

(i) U 1 1 21

2!

2 1

3!

2 ........

(ii) U 11

2

2 1

4

2 1

6

2 .........

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Solution:

(i) As standard series take the geometric series

V 1 1 21

2

2 1

22

2 ...........

Clearly,ur31

rand we know thatVconverges because its common

ratio is less than 1, and so U also converges.

(ii) U 11

2

2 1

4

2 1

6

2 ...........

Take V 1 1 2

series.

1

2

2 1

3

2 1

4

2 .........., the harmonic

Thenur1

2vrand sinceVdiverges, so doesU.

(b) Altering SeriesIn this particular case of an alternating series, such as:

11

2

2 1

3

1

4

2 .........

where the signs alternate, there is a very simple test for convergence.

Thus, let

S=u1u2+u

3u4+..........

allui.0 . Then if

limun10 andu

n3u

n1for alln4N

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(Na certain finite number) thenthe series converges.

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(c) The Ratio TestThis useful test, the ratio test, deals with the positive values of the ratio ofsuccessive terms.

For the series

S 1 u1

2 u2

2 u3

2.......2un

2 ........

letlimn&

un21

un

=L

Then the series is Divergent ifL>1

Convergent ifL

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lim n& 1 n2 1 051

Therefore,the series converges.

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(ii) 1 22

1

2 22

2!

2 23

3!

2 24

4!

2 25

5!

n

2.........we haveun1

n!,

so

limn&

un21

un

1 limn&&

2n21n!

n21#!2n

1 lim2n&&

1

n21

1 051

Therefore, the series converges.

(iii) For the series 1 2

haveun1

n2,

122

2 132

2 142

2 152

2 .......... we

so

limun21

1lim

n&n

n&

1

n

n2

11

Therefore, in this case the ratio test is inconclusive. However, it ispossible to extend the method used for the harmonic series to provethat the above series converges and in fact, more generally, the series

1 21

2p

2 1

3p

2 1

4p

2 1

5p

2 ........

Converges ifp>1 and diverges ifp31. This series provides a good

standard series for use in a comparison test.

6.2 SEQUENCES

It is hoped that by now your students are able to understand sequences andseries. This will help them a lot in grasping the subsequent sections. Now we willmove on to arithmetic and geometric sequences.

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6.2.1 Ait/$etic%&'Geo$eticSe(!e&ces

The two simplest sequences to work with are arithmetic and geometric

sequences.

Anarithmetic sequence goes from one term to the next by always adding (orsubtracting) the same value. For instance, 1, 5, 9, 13, 17,... and 8, 5, 2, -1, -4,... is anarithmetic sequence, since you add 4 and subtract 3, respectively, in each step.

Ageometric sequence goes from one term to the next by always multiplying (or

dividing) by the same value. So 1, 3, 9, 27, 81,... and 64, 32, 16, 8, 4, 2,.1,2.. is a

geometric sequence , since we multiply by 3 and divide by 2, respectively, in each

step.

The number added (or subtracted) at each stage of an arithmetic sequence iscalled the common differenced, because if we subtract (find the difference of)successive terms, we will always get this common value.

d=a2 a

1=a

3 a

2

The number multiplied (or divided) at each stage of a geometric sequence iscalled the common ratior, because if we divide (find the ratio of) successive

terms, we will always get this common value.

r=a1

=a2

Consider the following examples:

Example 6.11:Find the common difference and the next term of the following sequence:

1, 4, 7, 10, 13, 16, ...

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Solution:To find the common difference, we have to subtract a pair of terms. It does notmatter which pair we choose, as long as they are successive terms:

16 13=313 10=310 7 =3 7 4=3 4 1=3

The difference is always 3, sod=3. Then, the next term is 16+3=19.

Example 6.12:

Find the common ratio and the sixth term of the following sequence: 2, 4, 8, 16,32,

Solution:

r141

81

161

3212

The sixth term is,T61ar

n1

12#2#6$1 12#25

164

For arithmetic sequences, the common difference isd, and the first term a1is

often referred to simply as a. Since you obtain the next term by adding thecommon difference, the value ofa

2is just a+d.

The third term isa3=(a+d)+d=a+2d.

The fourth term isa4=(a+2d)+d=a+3d.

Following this pattern, then-th term will beTn=an=a+(n1)d.

For geometric sequences, the common ratio isr, and the first term a1is often

referred to simply as a. Since you get the next term by multiplying the commonratio, the value ofa

2is justar.

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The third term isa3=r(ar)=ar

2

The fourth term isa4=rar2 #1ar3

Following this pattern, then-th term will have the formTn1a

n1ar

n1.

Example 6.13:Find the eighth term and then-th-term of the following sequence:

2, 1, 2, 4, 8, ...

Solution:The differences do not match: 2 1=1, but 4 2=2. So, this is not an arithmeticsequence.

On the other hand, the ratios are the same: 21=2, 42=2, 84=2. So this is

a geometric sequence with common ratior=2 and a=,2

.

To find the tenth andn-th terms, we can just plug it into the formula:

n1 (1) n1

n ,2

a81

,22#8$1 1

,227

1

,2128# 164

Example 6.14:Find then-th term and the first three terms of the arithmetic sequence witha

6=5

andd=2.

Solution:

Then-thterm is of the formTn1a

n=a+(n1)d. In this case, that gives us

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a6=a+(6 1)

2=5. Solving, we geta=

2.

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21r

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Sincea51ar

4, then:

41a(24)116a

641a

Thenan1

642n$1anda181

64217 #110240

6.2.2 Ait/$eticSeies

An arithmetic series is the sum of an arithmetic sequence. In an arithmetic series,the difference between one term and the next is constant. In other words, we justadd the same value each timeinfinitely. For example:

In general, we could write an arithmetic series like this:

{a,a+d,a+2d,a+3d, ...}

Where:

(a) ais the first term; and

(b) dis the difference between the terms (called the common difference).

Source: http://www.mathsisfun.com/algebra/sequences-sums-

arithmetic.html

We can write the arithmetic formula as follows:

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The formula for the firstnterms of an arithmetic sequence, starting withn=1, is:

!ai1n

+ a12an#i11

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The sum is, in effect,ntimes the averageof the first and last terms. This sum ofthe firstnterms is called then-th partial sum.

Example 6.17:

Find the 40thpartial sum ofa

n1

21n21

Solution:The 40th partial sum is the sum of the first forty terms. The first few terms are:

a11

21# 2 11

2

a21

22# 2112

a31

23# 211

2

The terms have a common differenced=2, so this is an arithmetic sequence.

Thelast term in the partial sum will bea401a

12 401#d# 13239

,

1

137.

Then,plugging this into the formula, the 35th partial sum is:

35(n) (35)(3 37) (35)(

40) i11 i ,2 1 n ,2,2

2 ,2,2

We can also derive another formula of the sum of the firstnterms of anarithmetic series as follows:

LetSnbe the sum of the firstnterms of an arithmetic series. The sum of an

arithmetic series from the first to the last term can be written as follows:

Sn1a2 a2d# 2 a22d# 2 a23d# 2.......2 a2 n1#d0 (1)

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TOPIC6 SEQUENCESANDSERIES !11.

And the sum of an arithmetic series from the last term to the first term can bewritten as follows:

Sn 1a2 n1#d2 /a2 n-2#d 2 /a2 n-3#d 2...........2a (2)

Add equation (1) to equation (2) above and we will obtain the sum of anarithmetic series as follows:

2Sn1 /2a2 n1#d0 2 2a2 n-1#d0 2............. until then

thterm

2Sn1n/2a2 n-1#d0

Sn1

2/2a2 n1#d0

Example 6.18:Find the sum of the first 20 terms of an arithmetic sequence: 1, 3, 5, 7, ..

Solution:The first term is a=1, and d=3 1=2, substitute into the formula

Sn 12

/2a2 n1#d0 12

/21# 2 201#(2)0110/22192# 0 14001S20

Example 6.19:If the first term of an arithmetic sequence is 2 and the last term i.e. the fifth term is10. Find the sum of the first 5 terms.

Solution:Use the formula:

i 11ai1

,2a12a

n# 1S

51a

12a

22a

32a

42a

5

ThusS51

,22210# 130

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Example 6.20:Given an arithmetic sequence 2, 5, 8, 11,.. Find the 15 terms and the sum of thefirst 15 terms.

Solution:We identify and geta=2,d=5 2=3,n=15.

Use the formulas below and substitute the respective values ofa,dandn.

an1a2 n1#d122 151# 3# 122(14)(3)122421441a

15

Sn1

2

/2a2 n1#d0 115

/22# 2 151# 3#0 115

/42420 115(23)13451S15

Example 6.21:Given an arithmetic sequence 10, 5, 0, 5, 10, .

Find the eighth term and the sum of the first ten terms.

Solution:We identify the value ofa,a=10, andd=5 10=5

Use respective formulas and substitute the values ofa,dandn:

an1a2(n1)(d)1102(81)(5)110351251a

8

Sn1

2/2a2 n1#d0 1

2/210# 2 101# 5#015/204501251S10

Example 6.22Given an arithmetic sequence 19, 22, 25, . Find:

(a) S4

(b) S100

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Solution:We havea=19,d=22 19=3,

(a) Sn1

2/2a2 n1#d0 1

2/219# 2 41#(3)012/3823(3)0 12(47)1941S4

(b) S100

1

100/2(19)2(1001)(3)0 150/38299(3)0 150(335)116750

6.2.3 Geo$eticSeies

A geometric series is the sum of a geometric sequence. A geometric sequence is asequence whereby every term is obtained when the previous term is multiplied

by the common ratio,r.

Take for instance, 2, 4, 8, 16 and 32. 2 is the first term, multiplied by 2 becomes 4;the second term, 4 multiplied by 2 becomes 8; the third term, 8 multiplied by 2

becomes 16; the fourth term, 16 multiplied by 2 becomes 32; the fifth and the lastterm.

The common ratio is obtained by the following formula: r1an21. Thus, from

n

the above example we can calculate the common ratioras follows:

4 8 16 32

2 4 8 16progression as follows:

Hence, in general we can write the geometric

a, ar,ar2,ar

3,...........,ar

n-1

Therefore thenth term of a geometric sequence isn1

n n

If we would like to find the 20th term of the example above then we justsubstitute the values ofa=2,r=2, andn=20 into the formula as follows:

T201a

201 2# 2#20$1 1 2#2

19

11048576

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We write down the geometric series from the first term to thenthterm as follows:

Sn1

a 2

ar2

ar

22

ar

32

...........2

ar

n-12

........ (1)

Then we multiply equation (1) by 2 to obtain equation (2) below:

rSn1 ar2ar

2

2 ar3

2 ar4

2...........2arn

2 ......... (2)

Subtract equation (1) with equation (2) and we obtain the following:

SnrSn =aar

n

Sn1 r# =a1r

n

# (3)

Sn=a

"1 r ##wherer1

Example 6.23:Find the sum of the first 10 terms of the geometric series below:

(a) 121

2

2 1

4

2 1

8

2 66.

(b) 4+8+16+32+..

Solution:

(a)1 2 1 4 1 8

11 1 2 1 4

2

S101

1 1 2#10 # 1023 1024 1023

5111 1 2# 1 2 512

512

(b)8 16 32

4 8 16

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S1 0

=a#112#0 #

=4#(-1023)

=(4)(1023)=4092

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Example 6.24:Given a geometric sequence 3, 9, 27, 81, . Find the fifth term and the sum of thefirst five terms.

Solution:

For the question, get:a=3,r191

271

8113, andn=5

The fifth termT51a

513# 3#5$1 13#3

4

13x811243

The sum of the first five terms=S513#

11$3# #13#

$22#13x1211363

Example 6.25:Given a geometric sequence 3, 9, 27, 81, .., 2187. How many terms are there inthe sequence?

Solution:

We obtaina=3,r191

271

8113 and the last term,T

n12187

Using the formula:Tn 1arn1 ;

218713# 3#n$1 ;

Tn 13# 3#n$1

2187 n1

3

takinglogon both sides of the equal sign:

log(729)1n1#log3# ;log729#

log3#

Hence, the last term isn=6+1=7. i.e. 2187 is the 7th term of the sequence.

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Example 6.26:

Given a geometric sequence: 1, (1.05),1.05#2,1.05#3,

Find the ninth term and the sum of the first nine terms.

Solution:We identify and get:

1.05#2

1.05#31.05

1.05#2

T911# 1.05#9$1 11.05#8 11.4475 (4 decimal places)

S911#

11$1.05# #1

0.551328216111.027 (3 decimal places)

Example 6.27:What is the smallest term that can be taken from the sequence 1, 3, 9, 27, . so thatits sum will be greater than 1000?

Solution:We identify and get:

a=1,r131

91

2713

1 1$3 #

41000

3n 1# 42000

; 3n 1#

41000

; 3n

42001

taking log on both sides of the equal sign:

nlog3# 4log2001# :log2001#

log(3)

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n>6.9191, thus the smallest term is 7.

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Example 6.28:The sum of the first three terms of a geometric sequence is 117 and the sum of the

next three terms is43. Find the third term of the geometric sequence.

Solution:We consider the problem as follows:

(T12T

22T

3)1117 2 T

42T

52T

6# 14

3

3

S

3

1 a

1-r#

1117 (1) ;S

61a

1- r6#111724

11

364

.. (2)

Divide equation (2) by equation (1):

1r6 # 364

31r3 # 117

12r3 #

3641

351

3 13 1

351 27

;

;

;

1r3 #12r3 #

3641r3 #351

3 364

351

r13

271

3

Now we find the first term,aand the third term,T3:

91

(1)

3 :Substituter1

3into equation (1): a

(1)1117

,3

a,

127

1117%2

3

; a,27

178

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a178%26

181

2The third term,T

31ar

2

181#,3

19

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6.2.) T/eS!$toI&fi&it0of%Co&"e+e&ceGeo$eticSeies

An infinite series is a series whereby the number of terms is unlimited anduncountable.

For example:

(a) 1+2+3+4+..

(b) 1 21

2

2 1

4

2 1

8

2 ...............

In a case of r

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nr1 r1 r1

Whennapproaches infinityn& #,r

& thereforeS&

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Hence, in this case whenS& , the sum to infinity of its geometric series does

not exist.

Let us consider the following examples.

Example 6.29:In the following geometric series, determine the sum to infinity if it exists.

(a) 8, 2,2,

(b) 2, 2.2, 2.42,

Solution:

(a) We identify and get,a=8,r181

4Since15r51 , thenS

exists.

S1

1

a

r

118

1

1102

(b) We identify and get, a=2,r12.2

11.1

Sincer>1, thenSdoes not exist.

Example 6.30:The sum to infinity of a geometric series is 8. If the second term of this series is 2,find the sum of the first six terms.

Solution:Given thatT

2

ar

=2

=2........................................ (1)

and S

a

1r

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=8 =8........................................ (2)

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2# = 1#: a

1-r

= ar1 8

2

a

1r

% 1

ar

14

1

rr2

14

11 4r4r2

4r24r2110

2r1#2 10

Thus,r11

2

Substituting the value ofr12into (1), we have:

(1)

,2

a=4

> S6

a1r6 #

1r

6

4;11,2

20140826053327_10 HBMT4203 Topic 6(1)

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