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AN INTRODUCTION TO SEQUENCE AND

AN INTRODUCTION TO SEQUENCES AND SERIES

By

NEETHU KRISHNAN

(B.Sc , Mathematics)

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Preface

The book “ An Introduction to sequences and series” is intended for the secondary students andteachers in Kerala syllabus. In this book all the topic have been deal with in a simple and lucidmanner. A sufficiently large number of problems have been solved. By studying this book , thestudent is expected to understand the concept of arithmetic progression, geometric progression,harmonic progression, practical situations. To do more problems involving the arithmetic,geometric ,harmonic progression.

Suggestion for the further improvement of this book will be highlyappreciated.

Neethu krishnan.

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CONTENTS

Title Page No:

Preface

Chapter 1. Sequence and progressions 4-5

Chapter 2. Arithmetic progression 6 -11

Chapter 3. Geometric progression 12-13

Chapter 4. Arithmetico-Geometric Sequance 14-16

Chapter 5. Harmonic progression 17-25

REFERENCE 26

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Chapter-1

SEQUNCES AND PROGRESSION

SEQUENCES

Definition-1

A set of numbers occurring in a definite order or formed according to some definiterule is called a sequence.

Thus any set of numbers a1, a2, a3 ………,an,… such that to each positive integer n ,there corresponds a number an is a sequence. The numbers a1, a2, a3 ….………,an,… are called the elements or members of the sequence.

NOTE:

if N = { 1,2,3,…………n……..} and

Nn = {1,2,3,………..,n}, a sequence in a non-empty set x is defined as a maps fromN or Nn→ X i.e., s : N → X or s : Nn → X.

If X = R , the sequence s is called a real sequence .

If X = C , the sequence s is called a complex sequence.

A sequence is said to be finite or infinite according as the number of element in it is finite or infinite i.e., according as dom s =Nn or dom s =N.

If s is a sequence, then for any kN OR Nn, s(k) X is usually denoted by ak.

Thus s(k) = ak X where X is either R or C.

Then a1, a2, a3 ………,an (or a1, a2, a3 ………,an) determine the function s or the sequence s.

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In symbols {an}nk+1 denotes the finite sequence a1, a2, a3 ………,an and {an}n=1 or

simply {an} denotes the infinite sequence a1, a2, a3 ………,an ,……..

Here ak is called the kth term of the sequence. Incidentally, if k ,ak-1 is the

preceding term of ak and ak=1 is the succeeding term for k

Also a1 is called the first term. In the case of the finite sequence a1, a2, a3 ………,an

;an is called the last term. ak, ak=1 are called consecutive term.

Sequences following certain patterns are more often called progressions.

SERIES

Definition-2

A sequence fof numbers, connected together by the sign of addition (+), is called a series.

Thus a1+a2+ ………..

1+4+7+…………..

2+8+32+……are series.

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Chapter-2

ARITHMETIC PROGRESSION

Definition -1

A finite or infinite sequence a1, a2, ……….. anor a1, a2, ……….. an, ……… is called an arithmetic progression (abbreviate as A.P) if ak –a k-1=d constant independent ofk , for k =2,3,………n or k = 2,3,…..n,……..as the case may be.

Here note that the difference between any two consecutive term is a constant, denoted by d. this constant is called the common difference (abbreviated as C.D.)ofthe arithmetic progression.

Remark.

If each term of a sequence in A.P. be (i) increased , (ii) decreased, (iii) multiplied or (iv) divided by the same quantity (except 0 in case of division ) the resulting sequence is also in A.P.

Definition-2

The n th term of an A.P.

In an A.P. whose first term is a1 and common difference is d, the formula for the

n th term is

an = a1 + (n-1)d.

Example 1. Find the nth term of sequence 5, 2, -1, -4, -7………

Solution. Since the difference between any two consecutive terms of the sequence is same, namely -3, it is clearly in A.P.

Here a1 =5 and d = -3.

nth term =an

= a1 + (n-1) d = 5+ (n-1)(-3)

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=5- 3n +3=8 - 3n.

Example 2. The 10th term of an A.P .is 73 and the 20th term is 43 . find the 44th term .

Solution. Let a1 be the first term and d be the C.D.

Then a10 = a1 + 9d = 73. ……..(1)

and a20 = a1+ 19d = 43. ……..(2)

solving (1) and (2), we get a1 = 100 and d = -3.

44th term = a1 + 43d =100 +43 * (-3)

= -29.

Example 3. A sequence {an} is given by an = n2 – 1, nN. Show that it is not an A.P

Solution. We have an =n2-1. an+1 = (n+1)2 -1 = n2 =2n.

an+1 – an =(n2=2n) - (n2-1) =2n +1, which is not independent of n.

Hence {an} is not an A.P

Definition-3.

The sum of n term of an A.P.

Let Sn denote the first n terms an A.P.in which the first term is a and the common difference is d.

Then,

Sn = n/2 [1st term + nth term].

Sn = n/2 [2a + (n-1) d].

Example1: Find the sum to n term of sequence {an}, where an =5-6n, nN.

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Solution. We have an =5-6n. an+1 =5-6 (n+1)

= -6n -1.

an+1-an =-6n-1 –(5-6n) =-6, a constant independent of n.

Hence {an} is an A.P.

A1 =1 st term =5-6*1

= -1.

sn = sum to n terms =n2[1st term + nth term]

=n/2[-1=5-6n]

=n/2(4-6n)

=n (2-3n),

Example2. Find the sum of all natural number between 100 and 1000 which are multiples of 5.

Solution. The natural numbers between 100 and 1000 which are multiples of 5 form an A.P. namely 105, 110, 115,…….,995, where a = 105 and d =5. Suppose there are n terms in this finite sequence so that 995 is its nth term.

105 + (n-1) 5 = 995.

solving n= 179.

required sum = 179/2 [1st term + 179th term]

= 179/2 (105 +995)

= 179/2 * 1100

= 98450.

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ARITHMETIC MEAN (A.M.)

Definition: when three quantities are in A.P., the middle one is called the

Arithmetic mean (addreviated as A.M.) between the other two.

Let x be the A.M. between a and b. then a, x, b are in A.P.

x-a =b-x i.e., 2x =a + d

A.M. = a + b2.

ARITHMETICMEANS (A.M.S)

Definition: In an Arithmetic progression of n terms, the terms between the first and the last are called the Arithmetic means (abbreviated as A.M.S) between them.

To insert n arithmetic means between a and

Let a2, a3, ………….an+1 be the n arithmetic mean between a and b.

The n arithmetic means between a and b are

(na + b)/(n+1),[ (n-1)a+2b]/(n+1), (n-2)( a+3b)/(n+1),………., (a + nb)/(n+1)

Example 1. Find three numbers in A.P.whose sum is 21 and whose product is 280.

Solution . let the numbers be a –d, a, a+d.

Then (a-d) + a + (a+d) =21, i.e., 3a =21.

a =7.

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Also (a-d) a (a +d) = 280 i.e., a(a2 –d2 ) = 280.

Substituting for a in the above, we get

7(49 –d2) =280. i.e., d2 = 9.

d= +- 3.

Hence the required number numbers are

7-3, 7,7+3 or 7-(-3),7=(-3).

i.e., 4, 7, 10.

Exercise

1. Find the 20th of term of an A.P. whose first term is 5 and common difference

is 2.

2. Find the 8th and 20th terms of the sequence 11, 14, 17,……….

3. Find the nth term of the sequence 12, 7, 2, -3, -8,……………

4. Determine k so that k+2 , 4k-6 and 3k-2 are the three consecutive terms

of an A.P.

5. The third term of an A.P. is 25 and the tenth term is -3. Find the first term and the common difference.

6. The 4th and 15th term of an A.P. are11 and 55 respectively. Find its 20th term.

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7. The 7th term of an A.P.is the 15th term is 74. Determine the 1st term and the 40th term.

8. Which term in the A.P. 68, 64, 60, ……..is -8 ?

9. A sequence {an} is given by the formula an =10-3n.prove that it is an A.P.

10. Find the sum to

(i) 15 term of the sequence 3, 7, 11, …..

(ii) 20 term of the A.P. 10, 7, 4, …………

(iii) 81 term of the A.P. -1 ,1/4, 3/2,……….

(iv) The nth term of an A.P.is4n -1. Show that it is an A.P. and find the sum of n terms.

11. Determine the sum of 32 terms of an A.P. whose third term is 1and the 6th term is -11.

12. Determine the sum of first thirty five term of an arithmetic progression if

a2 = 2 and a7 =22.

13. Find the sum of all natural numbers with two digits.

14. Find the sum of all natural numbers between 99and 1001 which are multiples of 5.

15. Find the sum of first hundred even natural numbers which are divisible by 5.

16. Insert three A.M.s between3 and 19.

17. Insert seven A.M.s between -5 and 11.

18. Find four number s in A.P. whose sum is 20 and sum of whose squares is 120.

19. The sum of three numbers in A.P. is -3 and their product is 8. Find the numbers.

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20. Find the sum of all natural number between 50 and 500 which are divisible by 7.

Chapter-3

GEOMETRIC PROGRESSION (G.P.)

Definition-1

A finite or infinite sequence a1, a2, ………, an or a1, a2, ……, an, ……. Where none of the an’s is zero is called a geometric progression (abbreviated as G.P ) if

ak+1/ak=r, aconstand ( i.e., independent of k ), for k=1, 2, ………,n-1

or k=1, 2, 3,………,n,……..as the case may be .

the constant ratio r is called the common ratio (abbreviated as C.R)of the G.P

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Definition -2

The nth term of a G.P.

= a1 r n-1

thus , if a1 =a is the first term of a G.P. with common ratio r

Problem1. Find the 5th and 9th term of the sequence 5, 10, 20,...................

Solution. The given sequence is a G.P. where a =5 , r=2.

an = a r n-1

= 5.2 n-1.

a5 = 5.2 5-1

= 5. 24

a5 = 5*16 =80.

and a9 =5. 2 9-1

=5.28

= 5*256

a9 =1280.

Problem-2. Which term of the G.P. 3, 6, 12, ....is 768 ?

Solution. Here a =3 and r =2.

Let 768 be the nth term of the G.P. i.e., an =768.

But an = a r n-1

= 3.2 n-1.

3.2n-1 = 768 . i.e.., 2n-1 = 256 = 28.

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n-1 = 8 or n = 9 .

Problem -3. The first term of G.P. is 1. The sum of third and fifth terms is 90 .

Find the common ratio of the G.P.

Solution . let a, ar ar2 ,............. be the given G.P.

Then first term =a, third term =ar2, fifth term =ar4.

Given a= 1 and ar2 +ar4 =90.

1 (r2+r4) =90 or r4 +r2-90 =0.

i.e., (r2 + 10)(r2 -9) =0 r2 = -10 or r2 =9.

But r2 = -10 is inadmissible beause in this case r is imaginary.

r2 = 9 and r = +-3.

Chapter-4

Geometric mean (G.M)

Definition.1

Geometric mean (G.M.)

If three number are in G.P., the middle one is called the geometric mean

( abbreviated as G.M.) between the other two.

To find the G.M. between a and b

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Let x be the G.M. between a and b . then .by difinition, a,x,b are in G.P.

x=√ab.

Hence . G.M. between a and b =√ab.

Geometric means (G.MS.)

Definition .2

In a function sequence in G.M., the term between the first and the last are called the geometric means (abbreviated as G.M.) between them.

To insert n G.M. s between a and b

Let a2, a3,..............an+1 be an+2 =bis a sequence in G.P. with first term a and (n+2)th term b. Therefore, if r is the C.R. of the G.P.,

The n geometric means between a and b are

an/(n+1)b1/(n+1), a(n-1)/n+1)b2/(n+1) ,..................a1/(n+1) bn(n+1).

Problem1: Insert 3 geometric means between the number 1 and 256.

Solution. Let the numbers inserted be a2, a3, a4, so that 1, a2, a3, a4, 256 are in G.P.

Let r be the C.R. of the G.P.

Then 256=a5 = 1. r 5-1 = r4. i.e., r4 =(+-4)4. r =+-4.

When r = 4, a2 = 1 . r = 4, a3 =1. r2 = 16 , a4 = 64.

When r = -4, a2 = - 4. a3 = 16,a4 = -64.

Thus 4, 16, 64. Or -4,16, - 64 are the G.Ms.

Problem .2: The arithmetic mean between two number is 34 and their geometric mean is 16. F ind the number.

Solution. Let the two numbers be a and b.then

A.M. =34. i.e., a +b/2 =34. a+b = 68. .................(1)

G.M. =16. i.e., √ab =16. .ab =256. .................(2)

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From (1), b = 68 –a.

Putting this value of b in (2),

a (68 –a) =256. i.e. a2 – 68a + 256 = 0.

(a -4) (a – 64) = 0. either a = 4 or 64.

Whena =4, b =68 -4 = 64.

When a = 64, b = 68 -64 =4.

Hence the required numbers are 4 and 64.

Problem3: Find three numbers in G.P. whose sum is 26 and product is 216.

Solution. Let the numbers in G.P. be a/r , a, ar.

Then a/r . a. Ar = 216. i.e., a3 = 216. a = 6.

Also a/r+a+ar =26. i.e., 6/r+6+6r =26.

6r2 – 20r +6 = 0. r = 1/3 or 3.

the required numbers are 2, 6,and 18.

Remark: In such problems if there are 4 numbers in G.P., it is some times convenient to take them as a/r3 ,a/r, ar, ar3.

Sum to infinity of a G.P.

The sum of the infinite series a+ ar + ar2 + .........is a/1-r, provided r 1.

Thus s ∞ =a/1-r if r 1.

Problem-4: find the sum to infinity of the G.P. -5/4, 5/16, -5/64 ..................

Solution. Here a= -5/4 and r = -1/4.; also r 1.

sum to infinity = a/1-r = -5/4/1-(-1/4) = -5/4/5/4 = - 1.

Problem 2 the first term of a G.P., is 2 and the sum to infinity is 6. Find the common ratio.

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Solution. We have a= 2. Let r be the common ratio.

Given s∞ = 6. i.e., a/1-r = 6. i.e., 2/1-r = 6.

6 (1-r) = 2 i.e., 6 – 6r = 2. r =4/6 = 2/3 .

Hence common ratio = 2/3.

Exercise1. Write down the 4th and 9th term of the sequence 256, 128, 64,.........

2. Find the 6th and 10th term of the G.P..12, 8, 16/3,................

3. The first term of a G.P. is 50 and fourth term is 1350. Find the 5th term.

4. The third term of G.P. is 4 and 7th term is 36. Find the 8th term.

5. The 4th , 7th and 10th term of a G.P. are a, b, c, respectively. Show that b2 =ac.

6. Which term of the G.P. 2, 2√2, 4,.................is 64 ?

7. Find the sum to n term of

(i) 3 +33 + 333+............... (iii) 7 + 77 + 777 +..............

(ii) 4 + 44 + 444+................ (iv) 9 + 99 + 999 +.............

Chapter-5

Arithmetico geometric sequence

Definition .1

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A sequence in which each term is the product of the corresponding term of an arithmetic and a geometic and a geometric progression is called an arithmetic geometric sequence ( briefyly e.g. sequence).

Arithmetico-geometric sequence is

a, (a +d )r, (a + 2d)r2,..........

nth term s of the A.G. sequence =[a + (n-1) d ] r n-1.

Sum to n term of an A.G. sequence

Sn = a/ 1-r + dr ( 1-r n-1)/(1-r) -[a+(n-1)d]rn/1-r

Sum of infinity of the A.G. series

The a.g series a + (a+d)r + (a+2d)r2+...............

Hence s =a/1-r + dr/(1-r)2. Provided r 1.

Theorem -1 Sum to n terms of special sequences

The sum of the first n natural numbers

The first n natural numbers 1, 2, 3,...........,n from a sequence in a.p., where a = 1 and a = 1 . therefore if Sn denotes their sum

The sum of the squares of the first n natural numbers

Let sn = 12 +2 2 + 32 +........................+ n2.

Sn = n (n+1) (2n +1)/6

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The sum of the cubes of the first n natural numbers.

Let sn = 13 + 23 +33 + ...................+ n3 .

Sn = { n(n+1)/2}2

Remark

we found that

13 + 23 + ..................+ n3 = { n(n+1)/2}2 = (1 + 2+ 3+ .................+ n)2 .

Hence the sum of the cubes of the first n natural numbers is the sequance of the sum of the first n natural numbers.

Problem find the sum of n terms of the series 12 +32 + 5 2+ ...........

Solution let tn denote the nth tern of the given series.

Then tn =[ 2 + (n-1) . 2]2 = ( 2n -1)2 = 4n2 – 4n +1.

sum of n terms =

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= 4* n(n+1) (2n +1)/6 -4* n(n+1)/2 +n

= 2n (n+1) (2n + 1) /3 - 2n (n +1)+n

= n/3 [2 (n+1) (2n +1) – 6 (n+1) +3]

= n/3 (4n2 +6n + 2 -6n -6 +3)

= n/3 (4n2 -1)

Problem sum to n terms of the series whose nth term is n(n+1) (n +4).

Solution let tn do note the nth term.

Then tn = n(n+1) (n+4)

=n3 +5n2 +4n.

sum of n terms =

= {n(n=1)/2}2 + 5 . n(n+1) (2n+1)/6 + 4. n(n+1) /2

= n(n=1)/12 [3n (n+1) + 10 (2n +1) +24]

= n(n=1)/12 ( 3n2 +23n+ 34).

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Exercise

Find the sum of n terms of the following series (problems 1 to 3 ):

1. 1 + 3x + 5x2+...................

2. 1 + 3.2 + 5.22 + 7.23+ ..............

3. 3.2 + 6.22 + 9.23 +................

Find the sum to infinity of the following series (problems 4 to 9):

4. 1 + 2a + 3a 2 + 4a 3 + ..................if a 1

5. 1+ 3/2 +5/22 + 7/23 +..........

6. 1- 3/2 +5/4 -7/8+..................

7. 2 +3/2 +1 +5/8 + ...................

8. 1- 2/3 + 1/3 – 4/27 +...................

9. 3 +5.1/4 +7.4/42 +................

10. If the sum to infinity of the series 3 +5r + 7r2 + .................is 4 8/9 , find r.

11. Find the sum of n term of the series (questions 12 to 19):

12. 22 + 42 + 62 + ....................

13. 22 + 52+ 82+ ........................

14. 3.7 + 7.12+11.17+....................

15. 13 + 33+ 53+ ..............................

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16. Show that 1.2 2 + 2.3 3 + .....................+ n(n+1)2/12.2 + 22.3 + ..................+ n2(n+1)

= 3n +5/3n +1.

Chapter-6

Harmonic progression (H.P)

Definition-1

A sequence a1, a2, ........................an ,.........is called a harmonic progreesion (abbreviated as H.P. ) if

1/a1 ,1/a2, 1/a3 ,..............,1/an ,....................

Is an arithmetic progression.

For example

½, ¼ ,1/6,............., 1/4,1, -1/2,-1/5,-1/8,................., 1/a, 1/a+d, 1/a+2d,...........

Are in H.P.., since

2, 4, 6, ........... 4. 1, -2 , -5 ,-8 ,......................., a, a+d, a+2d,.................

Are in A.P.thus corresponting to every H.P.there is an A.P. and vice versa.

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Hence problems in H.P.( except the summation) can generally besolved with reference to the corresponding A.P.

Harmonic mean (H.M.)

Definition -2

When three quantities are in H.P.,the middle one is called the harmonic mean

( abbreviated as H.M.) between the other two.

The H.M. between two numbers a and b

Let x be the H.M. between a and b. Then a, x, b are in H.P.

i.e., 1/a 1/x,1/b are in A.P.

x = 2ab/a+b.

Relation between A.M., G.M.,and H.M. of two real numbers

Theorm 1.

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If A,G,H,are the arithmetic, the geometric and the harmonic means between two positive real numbers a and b , then A,G,H, froma gometric progression.

Proof .

by definition,

A =a+b/2, G = √ab H =2ab/a+b.

A*H =a+b/2 *2ab/a+b

= ab = G2

Hence A, G, H, are in G.P.

Theorem 2

if A, G ,H are the arithmetic,the geometric and the harmonic means between two unequal positive real numbers a and b , then A , G, H are in descending

order of magnitude (i.e., A G H).

Proof. We have A = a+b/2, G = √ab , H = 2ab/a+b .

A - G =a +b /2 - √ab = ½ (a+b – 2 √ab) = ½ (√a -√b )2,

Which is always positive, since a and b are positive and unequal.

A > G ........(1)

Also G-H =√ab -2ab/a+b = √ ab / a+b (a+b -2√ab)

= √ab/a+b (√a-√b)2

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G > H. ..........(2)

From (1) and (2) , we observe that

A > G > H

Problem1 . The 3rd and 7th term of a H.P. are 12 and 2 respectively. Find the 13th

term.

Solution. Let a be the first term and d be the common difference of the corresponding A.P.

Then a+d = 1/12 and a+6d =1/2.

Solving a =-1/8 and d =5/48.

13th term of the A.P. = a+12d = -1/8 + 12 *5/48 =9/8

Hence 13th term of the H.P. = 8/9.

Problem -2 . The A.M. between two numbers is 27 and their H.M . is 12.find their G.M.

Solution. Let a and b be the given numbers .

Then A.M. = a+b/2 = 27

And H.M. = 2ab/a+b =12.

From (1) and (2). We get ab = 324.

Hence G.M. = √ab = √324 =18.

Problem -3

If the pth term of a sequence in H.P. is q and the qth term is p, then show that the

(pq)th term is 1, provided p q

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Solution. Let a be the term and d be the common difference of the corresponding A.P.

Then a+(p-1)d =1/q

And a+ (q-1)d = 1/p.

Solving, we get a =1/pq, d = 1/q ,

(pq)th term of the A.P. = a=(pq-1)d = 1/pq =+(pq-1)1/q =1.

Hence, the (pq)th term of the H.P.=1.

Problem-4 : if b+c , c+a, a+b are in H.P., show that a2,b2,c2 are in A.P.

Solution. Since b+c, c+a, a+b are in H.P., we have

1/b+c, 1/c+a,1/a+b

are in A.P. 2/c+a= 1/b+c+1/a+b.

multiplying both sides by (a+b) (b+c) (c+a),we get

2(a+b)(b+c) = (c+a)(a+b)+(b+c)(c+a).

2ab2 =a2+c2

i.e.,

Hence a2, b2,c2 are in A.P

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problems

1. Find the 8th term of the H.P. 1/2 , 1/5 , 1/8,.......................

2. Find the 25th term of the H.P. 1/4 , 1/7, 1/10,....................

3. Find the nth term of the H.P. 3 , 6/5 ,3/4 ,......................

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4. The 3rd term of a H.P. is -1 and the 8th term is 1/9 ; find the 6th term.

5. The 5th term of a H.P. is 1/15 and 1th term is 1/21 .find the 20th term.

6. Find the valu of x for which x+10, x+11,x+13 willbe in H.P.

7. The mth term of a H.P. is n and the nth term is m. Prove that the pth term is mn/p.

8. In a H.P., if pth term is qr and qth term is pr ,prove that the rth term is pq

9. If pth , qth and r th terms of a H.P. are a,bc respectivly , prove that

q-r /a + r-p/b +p-q/c =0.

10. If H.M. and A.M. of two numbers are 3 and 4 respectively, find the numbers.

REFERENCE

1. MATHEMATICS TEXT FOR PLUS ONE : C.P.JANARDHANAN PILLAY

$ V.J .VIJAYALAKSHMI

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