vNatural numbers are the product of human spirit. – DEDEKIND v

9.1 Introduction

In mathematics, the word, “sequence” is used in much the

same way as it is in ordinary English. When we say that a

collection of objects is listed in a sequence, we usually mean

that the collection is ordered in such a way that it has an

identified first member, second member, third member and

so on. For example, population of human beings or bacteria

at different times form a sequence. The amount of money

deposited in a bank, over a number of years form a sequence.

Depreciated values of certain commodity occur in a

sequence. Sequences have important applications in several

spheres of human activities.

Sequences, following specific patterns are called progressions. In previous class,

we have studied about arithmetic progression (A.P). In this Chapter, besides discussing

more about A.P.; arithmetic mean, geometric mean, relationship between A.M.

and G.M., special series in forms of sum to n terms of consecutive natural numbers,

sum to n terms of squares of natural numbers and sum to n terms of cubes of

natural numbers will also be studied.

9.2 Sequences

Let us consider the following examples:

Assume that there is a generation gap of 30 years, we are asked to find the

number of ancestors, i.e., parents, grandparents, great grandparents, etc. that a person

might have over 300 years.

Here, the total number of generations

=

300

10

30

=

Fibonacci

(1175-1250)

Chapter

SEQUENCES AND SERIES

9

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178 MATHEMATICS

The number of person’s ancestors for the first, second, third, …, tenth generations are

2, 4, 8, 16, 32, …, 1024. These numbers form what we call a sequence.

Consider the successive quotients that we obtain in the division of 10 by 3 at

different steps of division. In this process we get 3,3.3,3.33,3.333, ... and so on. These

quotients also form a sequence. The various numbers occurring in a sequence are

called its

terms. We denote the terms of a sequence by a

1

, a

2

, a

3

, …, a

n

, …, etc., the

subscripts denote the position of the term. The n

th

term is the number at the n

th

position

of the sequence and is denoted by a

n.

The

n

th

term is also called the general

term

of the

sequence.

Thus, the terms of the sequence of person’s ancestors mentioned above are:

a

1

= 2, a

2

= 4, a

3

= 8, …, a

10

= 1024.

Similarly, in the example of successive quotients

a

1

= 3, a

2

= 3.3, a

3

= 3.33, …, a

6

= 3.33333, etc.

A sequence containing finite number of terms is called a finite sequence. For

example, sequence of ancestors is a finite sequence since it contains 10 terms (a fixed

number).

A sequence is called infinite, if it is not a finite sequence. For example, the

sequence of successive quotients mentioned above is an infinite sequence, infinite in

the sense that it never ends.

Often, it is possible to express the rule, which yields the various terms of a sequence

in terms of algebraic formula. Consider for instance, the sequence of even natural

numbers 2, 4, 6, …

Here a

1

= 2 = 2 × 1 a

2

= 4 = 2 × 2

a

3

= 6 = 2 × 3 a

4

= 8 = 2 × 4

.... .... ....

.... .... ....

.... .... ....

.... .... ....

a

23

= 46 = 2 × 23, a

24

= 48 = 2 × 24, and so on.

In fact, we see that the n

th

term of this sequence can be written as a

n

=

2n,

where n is a natural number. Similarly, in the sequence of odd natural numbers 1,3,5, …,

the

n

th

term is given by the formula, a

n

= 2n – 1, where n is a natural number.

In some cases, an arrangement of numbers such as 1, 1, 2, 3, 5, 8,.. has no visible

pattern, but the sequence is generated by the recurrence relation given by

a

1

= a

2

= 1

a

3

= a

1

+ a

2

a

n

= a

n – 2

+ a

n – 1

, n > 2

This sequence is called Fibonacci sequence.

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SEQUENCES AND SERIES 179

In the sequence of primes 2,3,5,7,…, we find that there is no formula for the n

th

prime. Such sequence can only be described by verbal description.

In every sequence, we should not expect that its terms will necessarily be given

by a specific formula. However, we expect a theoretical scheme or a rule for generating

the terms

a

1

, a

2

,

a

3

,…,a

n

,… in succession.

In view of the above, a sequence can be regarded as a function whose domain

is the set of natural numbers or some subset of it. Sometimes, we use the functional

notation a(n) for a

n

.

9.3 Series

Let

a

1

, a

2

,

a

3

,…,a

n

, be a given sequence. Then, the expression

a

1

+

a

2

+

a

3

+,…+ a

n

+

...

is called the

series associated with the given sequence .The series is finite or infinite

according as the given sequence is finite or infinite. Series are often represented in

compact form, called sigma notation, using the Greek letter

∑

(sigma) as means of

indicating the summation involved. Thus, the series a

1

+ a

2

+ a

3

+

... + a

n

is abbreviated

as

1

n

k

k

a

=

∑

.

Remark When the series is used, it refers to the indicated sum not to the sum itself.

For example, 1 + 3 + 5 + 7 is a finite series with four terms. When we use the phrase

“sum of a series,” we will mean the number that results from adding the terms, the

sum of the series is 16.

We now consider some examples.

Example 1 Write the first three terms in each of the following sequences defined by

the following:

(i) a

n

= 2n + 5, (ii) a

n

=

3

4

n

−

.

Solution (i) Here a

n

= 2n + 5

Substituting n = 1, 2, 3, we get

a

1

= 2(1) + 5 = 7, a

2

= 9, a

3

= 11

Therefore, the required terms are 7, 9 and 11.

(ii) Here

a

n

=

3

4

n

−

. Thus,

1 2 3

1 3 1 1

0

4 2 4

a , a , a

−

= = − = − =

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180 MATHEMATICS

Hence, the first three terms are

1 1

2 4

– , –

and 0.

Example 2 What is the 20

th

term of the sequence defined by

a

n

= (n – 1) (2 – n) (3 + n) ?

Solution

Putting n = 20 , we obtain

a

20

= (20 – 1) (2 – 20) (3 + 20)

= 19 × (– 18) × (23) = – 7866.

Example 3

Let the sequence a

n

be defined as follows:

a

1

= 1, a

n

= a

n – 1

+ 2 for n ≥ 2.

Find first five terms and write corresponding series.

Solution

We have

a

1

= 1, a

2

= a

1

+ 2 = 1 + 2 = 3, a

3

= a

2

+ 2 = 3 + 2 = 5,

a

4

= a

3

+ 2 = 5 + 2 = 7, a

5

= a

4

+ 2 = 7 + 2 = 9.

Hence, the first five terms of the sequence are 1,3,5,7 and 9. The corresponding series

is 1 + 3 + 5 + 7 + 9 +...

EXERCISE 9.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose n

th

terms are:

1. a

n

= n (n + 2) 2. a

n

=

1

n

n

+

3. a

n

= 2

n

4.

a

n

=

2 3

6

n

−

5.

a

n

= (–1)

n–1

5

n+1

6. a

n

2

5

4

n

n

+

=

.

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n

th

terms are:

7. a

n

= 4n – 3; a

17

, a

24

8. a

n

=

2

7

;

2

n

n

a

9. a

n

= (–1)

n – 1

n

3

; a

9

10.

20

( – 2)

;

3

n

n n

a a

n

=

+

.

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SEQUENCES AND SERIES 181

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the

corresponding series:

11. a

1

= 3, a

n

= 3a

n – 1

+ 2 for all

n > 1 12. a

1

= – 1, a

n

=

1

n

a

n

−

, n ≥ 2

13. a

1

= a

2

= 2, a

n

= a

n – 1

–1, n > 2

14. The Fibonacci sequence is defined by

1 = a

1

= a

2

and a

n

= a

n – 1

+ a

n – 2

, n > 2.

Find

1

n

n

a

a

+

, for n = 1, 2, 3, 4, 5

9.4 Arithmetic Progression (A.P.)

Let us recall some formulae and properties studied earlier.

A sequence a

1

, a

2

, a

3

,…, a

n

,… is called arithmetic sequence or arithmetic

progression if a

n + 1

= a

n

+ d, n ∈ N, where a

1

is called the first term and the constant

term d is called the common difference of the A.P.

Let us consider an A.P. (in its standard form) with first term a and common

difference d, i.e., a, a + d, a + 2d, ...

Then the n

th

term (general term) of the A.P. is a

n

= a + (n – 1) d.

We can verify the following simple properties of an A.P. :

(i) If a constant is added to each term of an A.P., the resulting sequence is

also an A.P.

(ii) If a constant is subtracted from each term of an A.P., the resulting

sequence is also an A.P.

(iii) If each term of an A.P. is multiplied by a constant, then the resulting

sequence is also an A.P.

(iv) If each term of an A.P. is divided by a non-zero constant then the

resulting sequence is also an A.P.

Here, we shall use the following notations for an arithmetic progression:

a = the first term, l = the last term, d = common difference,

n = the number of terms.

S

n

= the sum to n terms of A.P.

Let a, a + d, a + 2d, …, a + (n – 1) d be an A.P. Then

l = a + (n – 1) d

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182 MATHEMATICS

[

]

S 2 ( 1)

2

n

n

a n d

= + −

We can also write,

[

]

S

2

n

n

a l

= +

Let us consider some examples.

Example 4 In an A.P. if m

th

term is n and the n

th

term is m, where m

≠

n, find the pth

term.

Solution We have a

m

= a + (m – 1) d = n, ... (1)

and a

n

= a + (n – 1) d = m ... (2)

Solving (1) and (2), we get

(m – n) d = n – m, or d = – 1, ... (3)

and a = n + m – 1 ... (4)

Therefore a

p

= a + (p – 1)d

= n + m – 1 + ( p – 1) (–1) = n + m

– p

Hence, the p

th

term is n + m – p.

Example 5 If the sum of n terms of an A.P. is

1

P ( – 1)Q

2

n n n+

, where P and Q

are constants, find the common difference.

Solution Let a

1

, a

2

, … a

n

be the given A.P. Then

S

n

= a

1

+ a

2

+ a

3

+...+ a

n–1

+ a

n

= nP +

1

2

n (n – 1) Q

Therefore S

1

= a

1

= P, S

2

= a

1

+ a

2

= 2P + Q

So that a

2

= S

2

– S

1

= P + Q

Hence, the common difference is given by d = a

2

– a

1

= (P + Q) – P = Q.

Example 6 The sum of n terms of two arithmetic progressions are in the ratio

(3n + 8) : (7n + 15). Find the ratio of their 12

th

terms.

Solution Let a

1

, a

2

and d

1

, d

2

be the first terms and common difference of the first

and second arithmetic progression, respectively. According to the given condition, we

have

3 8

7 15

Sum to termsof first A.P.

Sum to termsof second A.P.

n

n

n

n

+

+

=

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SEQUENCES AND SERIES 183

or

[ ]

[ ]

1 1

2 2

2 1

3 8

2

7 15

2 1

2

n

a ( n )d

n

n

n

a ( n )d

+ −

+

=

+

+ −

or

1 1

2 2

2 ( 1)

3 8

2 ( 1) 7 15

a n d n

a n d n

+ −

+

=

+ − +

... (1)

Now

th

1 1

th

2 2

11

12 term of first A.P.

11

12 term of second A.P

a d

a d

+

=

+

1 1

2 2

2 22

3 23 8

2 22 7 23 15

a d

a d

+

× +

=

+ × +

[By putting n = 23 in (1)]

Therefore

th

1 1

th

2 2

11

12 term of first A.P. 7

11 16

12 term of second A.P.

a d

a d

+

= =

+

Hence, the required ratio is 7 : 16.

Example 7 The income of a person is Rs. 3,00,000, in the first year and he receives an

increase of Rs.10,000 to his income per year for the next 19 years. Find the total

amount, he received in 20 years.

Solution Here, we have an A.P. with a = 3,00,000, d = 10,000, and n = 20.

Using the sum formula, we get,

20

20

S [600000 19 10000]

2

= + ×

= 10 (790000) = 79,00,000.

Hence, the person received Rs. 79,00,000 as the total amount at the end of 20 years.

9.4.1 Arithmetic mean Given two numbers a and b. We can insert a number A

between them so that a, A, b is an A.P. Such a number A is called the arithmetic mean

(A.M.) of the numbers a and b. Note that, in this case, we have

A – a = b – A, i.e., A =

2

a b

+

We may also interpret the A.M. between two numbers a and b as their

average

2

a b

+

. For example, the A.M. of two numbers 4 and 16 is 10. We have, thus

constructed an A.P. 4, 10, 16 by inserting a number 10 between 4 and 16. The natural

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