Sequences and Series Session MPTCP04. 1.Finite and infinite sequences 2.Arithmetic Progression...

Sequences and Series

Session MPTCP04

1. Finite and infinite sequences2. Arithmetic Progression (A.P.) -

definition, nth term3. Sum of n terms of an A.P.4. Arithmetic Mean (A.M.) and

insertion of n A.M.s between two given numbers.

5. Geometric Progression (G.P.) - definition, nth term

6. Sum of n terms of a G.P.

Session Objectives

Sequence – a Definition

A sequence is a function whose domain is the set N of natural numbers.

_I001

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

Finite and Infinite Sequences

Finite sequence :

a a1, a2, a3, . . ., an

Infinite sequence :

a a1, a2, a3, . . ., an, . . .

_I001

Series – a Definition

If

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

is a sequence,

_I001

the expression

a1+a2+a3+ . . . +an+ . . .

is called a series.

Arithmetic Progression

A sequence is called an arithmetic progression (A.P.) if the difference between any term and the previous term is constant.

The constant difference, generally denoted by d is called the common difference.

_I002

a1 = a

a2 = a+d

a3 = a+d+d = a+2d

a4 = a+d+d+d = a+3d

First termGeneral Term

an = a+d+d+d+... = a+(n-1)d

Is a Given Sequence an A.P.?

Algorithm to determine whether a

given sequence is an A.P. :

Step I Obtain general term an

Step II Determine an+1 by

replacing n by n+1 in the general term

Step III Find an+1-an. If this is independent of n, the given sequence is an A.P.

_I002

Problem Solving Tip

Choose Well!!!!

# Terms Common diff.

3 a-d, a, a+d d

4 a-3d, a-d, a+d, a+3d 2d

5 a-2d, a-d, a, a+d, a+2d d

6 a-5d, a-3d, a-d, a+d, a+3d, a+5d 2d

_I002

Illustrative problem

Q. If sum of three numbers in A.P. is 45, and the second number is thrice the first number, find the three numbers.

_I002

A. Let the numbers be a-d, a, a+d

Given that (a-d)+a+(a+d) = 45

3a = 45 a = 15

Also, a = 3(a-d) 3d = 30

d = 10

the three numbers are 5, 15, 25

Important Properties of A.P.s

If A a, a+d, . . ., a+(n-1)d

adding constant k to each term,

we get,

A’ a+k, a+d+k, . . ., a+(n-1)d+k

A’ is also an A.P. with the same common difference.

_I002


If A a, a+d, . . ., a+(n-1)d

multiplying each term by non-zero constant k,

A’ ak, ak+dk, . . ., ak+(n-1)dk

A’ is also an A.P. with common difference dk

_I002


ak+an-(k-1) = a1+an

k = 2, 3, 4, . . . (n-1)

Example :

Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

_I002


Example :

Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

a3+a8 = 22

_I002ak+an-(k-1) = a1+an

k = 2, 3, 4, . . . (n-1)

Important Properties of A.P.S

Example :

Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

a3+a8 = 22 = a5+a6 = 22

_I002ak+an-(k-1) = a1+an

k = 2, 3, 4, . . . (n-1)


Example :

Consider A 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

a3+a8 = 22 = a5+a6 = 22 = a1+a10 = 22

_I002ak+an-(k-1) = a1+an

k = 2, 3, 4, . . . (n-1)


a, b, c are in A.P. 2b = a+c _I002


A sequence is an A.P.

an = An+B, A, B are constants.

A is the common difference.

Proof :

an = a+(n-1)d

or, an = dn+(a-d)

or, an = An+B, where A is the common difference

_I002


If A a, a+d, . . ., a+(n-1)d

take every third term,

A’ a, a+3d, a+6d, . . . . . . . . . .

A’ is also an A.P.

_I002

Sum of n Terms of an A.P.

Sn = a1 +(a1+d)+ . . .+{a1+(n-2)d}+{a1+(n-1)d}

Also,

Sn = {a1+(n-1)d}+{a1+(n-2)d}+{a1+d}+. . .+a1

Adding,

2Sn = n{2a1+(n-1)d}

n 1n

S 2a n 1 d2

_I003

Sum of n Terms of an A.P.

n 1n

S 2a n 1 d2

This can also be written as :

n 1 1n

S a a n 1 d2

n 1 nn

S a a2

nn

S First Term Last Term2

_I003

Property of Sum of n Terms of an A.P.

A sequence is an A.P.

Sn = An2+Bn,

where A, B are constants.

2A is the common difference.

We know that, nn

S 2a n 1 d2

2n

d dS n a n

2 2Rearranging,

Or, Sn = An2+Bn.

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Arithmetic Mean

A is the A.M. of two numbers a and b

a, A, and b are in A.P.

A-a = b-A

2A = a+b

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a bA

2

Arithmetic Mean – a Definition

then A1, A2, A3, . . . , An are called arithmetic means (A.M.s) of a and b.

-4 -2 2 4A1 A2 A3 A4 A5

0-6

a

6

b

If n terms A1, A2, A3, . . . An are inserted between two numbers a and b such that a, A1, A2, A3, . . . , An, b form an A.P.,

_I004

Arithmetic Mean – Common Difference

Let n A.M.s be inserted between two numbers a and b

Let the common difference be d

Now b = a+(n+2-1)d = a+(n+1)d

-4 -2 2 4A1 A2 A3 A4 A5

0-6

a

6

b

The A.P. thus formed will have (n+2) terms.

_I004

mb a b a

d ; A a mn 1 n 1

Property of A.M.s

Let n A.M.s A1, A2, A3, . . ., An be inserted between a and b.

Then,

1 2 3 na b

A A A ... A nA n2

_I004

Illustrative Problem

Q. Insert 3 A.M.s between -4 and 3

A. Let the required A.M.s be A1,

A2 and A3.

Common difference d = 3 4 7

4 4

17 9

A 44 4

214 1

A 44 2

321 5

A 44 4

_I004

Geometric Progression

Consider a family where every

female of each generation has

exactly 2 daughters.It is then possible to determine the

number of females in each generation

if the generation number is known.

1st Generation 1 female

2nd Generation 2 females

3rd Generation 4 females

Such a progression is a Geometric Progression (G.P.)

_I005

Geometric Progression

A sequence is called a geometric progression (G.P.) if the ratio between any term and the previous term is constant.

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The constant ratio, generally denoted by r is called the common ratio.

a1 = a

a2 = ar

a3 = ar2

a4 = ar3

an = ar(n-1)

First term

General Term

Problem Solving Tip

Choose Well!!!!

# Terms Common ratio

3 a/r, a, ar r

4 a/r3, a/r, ar, ar3 r2

5 a/r2, a/r, a, ar, ar2 r

6 a/r5, a/r3, a/r, ar, ar3, ar5 r2

_I005

Important Properties of G.P.s

If G a, ar, ar2, . . ., arn-1

multiplying each term by non-zero constant k,

G’ ka, kar, kar2, . . ., karn-1

G’ is also a G.P. with the same common ratio.

_I005


If G a, ar, ar2, . . ., arn-1

taking reciprocal of each term,

G’ is also a G.P. with a reciprocal common ratio.

2 n 1

1 1 1 1G' , , , . . .,

a ar ar ar

_I005


If G a, ar, ar2, . . ., arn-1

raising each term to power k,

G’ ak, akrk, akr2k, . . ., akr(n-1)k

G’ is also a G.P. with common ratio rk.

_I005


akan-(k-1) = a1an

k = 2, 3, 4, . . . (n-1)

Example :

Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512

_I005


Example :

Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512

a3a8 = 512

_I005akan-(k-1) = a1an

k = 2, 3, 4, . . . (n-1)


Example :

Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512

a3a8 = 512 = a5a6 = 512


k = 2, 3, 4, . . . (n-1)


Example :

Consider G 1, 2, 4, 8, 16, 32, 64, 128, 256, 512

a3a8 = 512 = a5a6 = 512 = a1a10 = 512


k = 2, 3, 4, . . . (n-1)


a, b, c are in G.P. b2 = ac _I005


If G a, ar, ar2, . . ., arn-1

take every third term,

G’ a, ar3, ar6, . . .

G’ is also a G.P.

_I005


a1, a2, a3, . . . , an is a G.P. of positive terms

loga1, loga2, loga3, . . . logan is an A.P.

_I005

Sum of n Terms of a G.P.

Sn = a+ar+ar2+ar3+ . . .+ar(n-1) ………(i)

Multiplying by r, we get,

rSn = ar+ar2+ar3+ . . .+ar(n-1)+arn ……...(ii)

Subtracting (i) from (ii), (r-1)Sn = a(rn-1)

n

n

r 1S a

r 1

_I006

Class Exercise Q1.

Q. If log 2, log (2x-1) and log (2x+3) are in A.P., find x.

_I002

Class Exercise Q1.Q. If log 2, log (2x-1) and log (2x+3) are in

A.P., find x. _I002A. Given that

log(2x-1)-log2 = log(2x+3)-log(2x-1)

x x

x

2 1 2 3log log

2 2 1

2x x 1 x 12 2 1 2 6 2x x2 4.2 5 0

x x2 5 2 1 0

x x2 5 2 cannot be negative

2log5

x log 5log2

Class Exercise Q2.

Q. Show that there is no infinite A.P. which consists only of distinct primes.

_I002

Class Exercise Q2.

_I002A. Let, if possible, there be an A.P.

consisting only of distinct primes : a1, a2, a3, . . ., an, . . .

an = a1+(n-1)d

Q. Show that there is no infinite A.P. which consists only of distinct primes.

1a 1 1 1a a a 1 1 d

1a 1 1a a (1 d)

Thus, (a1+1)th term is a multiple of a1.

Thus, no such A.P. is possible.

Q.E.D.

Class Exercise Q3.

_I003Q. , where Sn denotes

the sum of the first n terms of an A.P., then common difference is :

(a) P+Q (b) 2P+3Q

(c) 2Q (d) Q

(J.E.E. West Bengal 1994)

nn

S nP n 1 Q2

Class Exercise Q3.

_I003Q. , where Sn denotes

the sum of the first n terms of an A.P., then common difference is :

(a) P+Q (b) 2P+3Q

(c) 2Q (d) Q

(J.E.E. West Bengal 1994)

nn

S nP n 1 Q2

A. an = Sn - Sn-1

nn 1n

a nP n 1 Q n 1 P n 2 Q2 2

na P n 1 Q

n n 1d a a

d P n 1 Q P n 2 Q d Q Ans : (d).

Class Exercise Q4.

_I003Q. If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

Class Exercise Q4.

_I003Q. If 12th term of an A.P. is -13 and the sum

of the first four terms is 24, what is the sum of first 10 terms?

A. Given that,

a12 = a1+11d = -13 . . . (i)

S4 = 2(2a1+3d) = 24 . . . (ii)

Solving (i) and (ii) simultaneously, we get,

a1 = 9, d = -2

S10 = 5(2a1+9d) = 5(18-18) = 0

Class Exercise Q5.

_I004Q. Find the value of n so that

be an A.M. between a and b (a, b are positive).

n 1 n 1

n n

a b

a b

Class Exercise Q5.

A. Given that,



n 1 n 1

n n

a b

a b

n 1 n 1

n n

a b a b

2a b

n 1 n 1 n na b a b ab

Dividing throughout by bn+1, we get,n 1 n

a a a1

b b b

na a a

1 1b b b

Class Exercise Q5.



n 1 n 1

n n

a b

a b

na a a

1 a b 1 1 0b b b

na a a1 1

b b b

n = 0

Class Exercise Q6.

_I004Q. 53 A.M.s are inserted between 2 and 98. Find the 27th A.M.

Class Exercise Q6.

_I004Q. 53 A.M.s are inserted between 2 and 98.

Find the 27th A.M.

A. Common differenceb a 98 2 48

n 1 53 1 27

2748

A a 27d 2 27 5027

Class Exercise Q7.

_I005Q. If the 3rd term of a G.P. is 4, what is the product of the first five terms?

Class Exercise Q7.

_I005Q. If the 3rd term of a G.P. is 4, what is the

product of the first five terms?

A. Let the first 5 terms of the G.P. be :

22

a a, , a, ar, arrr

Required product = a5

= (a3)5

=45

Class Exercise Q8.

_I005Q. If the 4th, 7th, 10th term of a G.P. are p, q, r respectively, then

(a) p2 = q2+r2 (b) q2 = pr

(c) p2 = qr (d) pqr+pq+q = 0

(M.N.R. 1995)

Class Exercise Q8.

_I005Q. If the 4th, 7th, 10th term of a G.P. are p, q, r

respectively, then

(a) p2 = q2+r2 (b) q2 = pr

(c) p2 = qr (d) pqr+pq+q = 0

(M.N.R. 1995)

A. Let the first term of the G.P. be and common ratio be .

Ans : (b)

p = 3, q = 6, r = 9

Now, pr = 212

= (6)2

= q2

Class Exercise Q9.

_I006Q. Find the sum to n terms of the sequence 6, 66, 666, . . .

Class Exercise Q9.

_I006Q. Find the sum to n terms of the sequence

6, 66, 666, . . .

A. Sn = (6+66+666+ . . .[n terms])

n6

S 9 99 999 . . . n terms9

n6

S 10 1 100 1 1000 1 . . . n terms9

2 3n

6S 10 10 10 . . . n terms n

9

n

n

10 16S 10 n

9 10 1

nn

6 10S 10 1 n

9 9

Class Exercise Q10.

_I006Q. How many terms of the G.P.

are needed to give the sum ?

2 1 1 3, , , , . . .

9 3 2 4

55

72

Class Exercise Q10.



2 1 1 3, , , , . . .

9 3 2 4

55

72

A. Common ratio1

332 29

Let the required number of terms be n.

n

n

n

31

55 2 2 2 32S 1

372 9 9 5 212

Class Exercise Q10.



2 1 1 3, , , , . . .

9 3 2 4

55

72n55 2 2 3

172 9 5 2

n

3 55 9 5 2431

2 72 2 2 32

n 53 3

2 2

n = 5

Thank you

Sequences and Series Session MPTCP04. 1.Finite and infinite sequences 2.Arithmetic Progression...

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