NCERT Solutions for Class 11 Maths Chapter 9 – Sequences ...

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NCERT Solutions for Class 11

Maths

Chapter 9 – Sequences and Series

Exercise 9.1

1. Write the first five terms of the sequences whose thn term is n

a n n 2 .

Ans:

The given equation is na n n 2 .

Substitute n 1 in the equation.

1a 1 1 2

1a 3

Similarly, substitute n 2,3,4 and 5 in the equation.

2a 2 2 2

2a 8

3a 3 3 2

3a 15

4a 4 4 2

4a 24

5a 5 5 2

5a 35

Therefore, the first five terms of na n n 2 are 3,8,15,24 and 35 .

2. Write the first five terms of the sequences whose th

n term is n

na

n 1

.

Ans:The given equation is n

na

n 1

.


1

1a

1 1

1

1a

2

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2

2a

2 1

2

2a

3

3

3a

3 1

3

3a

4

4

4a

4 1

4

4a

5

5

5a

5 1

5

5a

6

Therefore, the first five terms of n

na

n 1

is

1 2 3 4, , ,

2 3 4 5 and

5

6 .


n term is n

na 2 .


na 2 .

Substitute n 1 in the equation. 1

1a 2

1a 2

Similarly, substitute n 2,3,4 and 5 in the equation. 2

2a 2

2a 4 3

3a 2

3a 8 4

4a 2

4a 16 5

5a 2



5a 32


na 2 is 2,4,8,16 and 32 .

4. Write the first five terms of the sequences whose thn term is

n

2n 3a

6

.


2n 3a

6

.


1

2 1 3a

6

1

1a

6


2

2 2 3a

6

2

1a

6

3

2 3 3a

6

3

3 1a

6 2

4

2 4 3a

6

4

5a

6

5

2 5 3a

6

5

7a

6


2n 3a

6

is

1 1 1 5, , ,

6 6 2 6 and

7

6 .


n term is n 1 n 1

na 1 5

.



Ans:The given equation is n 1 n 1

na 1 5 .


1 1 1 1

1a 1 5

2

1a 5 25


2 1 2 1

2a 1 5 3

2a 5 125

3 1 3 1

3a 1 5

4

3a 5 625

4 1 4 1

4a 1 5 5

4a 5 3125

5 1 5 1

5a 1 5

6

5a 5 15625

Therefore, the first five terms of n 1 n 1

na 1 5 is 25, 125,625, 3125 and 15625


n term is 2

n

n 5a n

4

.

Ans:The given equation is 2

n

n 5a n

4

.


1

1 5a 1

4

1

6 3a

4 2

Similarly, substitute n 2,3,4 and 5 in the equation. 2

2

2 5a 2

4

2

18 9a

4 2

2

3

3 5a 3

4



3

42 21a

4 2

2

4

4 5a 4

4

4

84a 21

4

2

5

5 5a 5

4

5

150 75a

4 2

Therefore, the first five terms of 2

n

n 5a n

4

is

3 9 21, , ,21

2 2 2 and

75

2 .

7. Find the th17 and th

24 term in the following sequence whose thn term is

na 4n 3 .

Ans:The given equation is na 4n 3 .


17a 4 17 3

17a 65

Similarly, substitute n 24 in the equation.

24a 4 24 3

24a 93

Therefore, the th17 and th24 term of na 4n 3 is 65 and 93 respectively.

8. Find the th

7 term in the following sequence whose th

n term is 2

n n

na

2 .

Ans:The given equation is 2

n n

na

2 .


7 7

7a

2

7

49a

128



Therefore, the th7 term of 2

n n

na

2 is

49

128.

9. Find the th9 term in the following sequence whose th

n term is n 1 3

na 1 n

.

Ans:The given equation is n 1 3

na 1 n

.


9 1 3

9a 1 9

9a 729

Therefore, the th9 term of n 1 3

na 1 n

is 729 .

10. Find the th20 term in the following sequence whose

thn term is

n

n n 2a

n 3

Ans:The given equation is

n

n n 2a

n 3

.


20

20 20 2a

20 3

20

360a

23

Therefore, the th20 term of

n

n n 2a

n 3

is

360

23.

11. Write the first five terms of the following sequence and obtain the

corresponding series: 1

a 3 , n n 1

a 3a 2

for all n 1 .

Ans:The given equation is n n 1a 3a 2 where

1a 3 and n 1 .

Substitute n 2 and 1a 3 in the equation.

2 2 1a 3a 2 3 3 2

2a 11

Similarly, substitute n 3,4 and 5 in the equation.

3 3 1a 3a 2 3 11 2

3a 35

4 4 1a 3a 2 3 35 2



4a 107

5 5 1a 3a 2 3 107 2

5a 323

Therefore, the first five terms of n n 1a 3a 2 are 3,11,35,107 and 323 .

The corresponding series obtained from the sequence is 3 11 35 107 323 ...


corresponding series: 1

a 1 , n 1n

aa

n

, n 2 .

Ans:The given equation is n 1n

aa

n

where 1a 1 and n 2 .


2 12

a 1a

2 2

2

1a

2


3 13

1a 2a3 3

3

1a

6

4 14

1a 6a4 4

4

1a

24

5 15

1a 24a5 5

5

1a

120

Therefore, the first five terms of n 1n

aa

n

is 1 1 1

1, , ,2 6 24

and 1

120 .



The corresponding series obtained from the sequence is

1 1 1 1

1 ...2 6 24 120


corresponding series: 1 2

a a 2 , n n 1

a a 1

, n 2 .

Ans:The given equation is n n 1a a 1 where

1 2a a 2 and n 2 .


3 3 1a a 1 2 1

3a 1

Similarly, substitute n 4 and 5 in the equation.

4 4 1a a 1 1 1

4a 0

5 5 1a a 1 0 1

5a 1

Therefore, the first five terms of n n 1a a 1 is 2,2,1,0 and 1 .

The corresponding series obtained from the sequence is 2 2 1 0 1 ...

14. The Fibonacci sequence is defined by 1 2

1 a a , n n 1 n 2

a a a

, n 2 . Find

n 1

n

a

a

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

Ans:The given equation is n n 1 n 2a a a where

1 21 a a and n 2 .

Substitute n 3 and 1 21 a a in the equation.

3 3 1 3 2a a a 1 1

3a 2


4 4 1 4 2a a a 2 1

4a 3

5 5 1 5 2a a a 3 2

5a 5

6 6 1 6 2a a a 5 3

6a 8



Substitute the values of 1a and

2a in the expression n 1

n

a

a

for n 1 .

1 1

1

a 11

a 1

Similarly, when n 2 ,

2 1

2

a 22

a 1

When n 3 ,

3 1

3

a 3

a 2

When n 4 ,

4 1

4

a 5

a 3

When n 5 ,

5 1

5

a 8

a 5

Therefore, the value of n 1

n

a

a

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

1,2, ,2 3

and 8

5 respectively.

Exercise 9.2

1. Find the sum of odd integers from 1 to 2001 .

Ans:1,3,5,...,1999,2001 are the odd integers from 1 to 2001 . An A.P. is formed by

this sequence.

The first term of the A.P. is a 1 and the common difference is d 2 .

The thn term of the A.P. is given by the equation na a n 1 d .

Therefore, a n 1 d 2001

Substitute a 1 and d 2 in the equation.

1 n 1 2 2001

2n 1 2001

2001 1n

2

n 1001



The sum of first n terms of an arithmetic progression is given by the equation

n

nS 2a n 1 d

2 .

Substitute the values of n , a and d in the equation.

n

1001S 2 1 1001 1 2

2

1001

2 1000 22

10012002

2

1001 1001

1002001 Therefore, 1002001 is the sum of the odd integers from 1 to 2001 .

2. Find the sum of all natural numbers lying between 100 and 1000 , which are

multiples of 5 .

Ans:105,110,...,990,995 are the natural numbers lying between 100 and 1000 , which

are multiples of 5 . An A.P. is formed by this sequence.

The first term of the A.P. is a 105 and the common difference is d 5 .




105 n 1 5 995

5n 100 995

995 100n

5

n 179 The sum of first n terms of an arithmetic progression is given by the equation

n

nS 2a n 1 d

2 .


n

179S 2 105 179 1 5

2

179

2 105 178 52



179 105 89 5

179 550

98450 Therefore, 98450 is the sum of the natural numbers lying between 100 and 1000 ,

which are multiples of 5 .

3. In an A.P., the first term is 2 and the sum of the first five terms is one-fourth

of the next five terms. Show that th20 term is 112 .

Ans:The first term of the A.P. is 2 and let the common difference of the A.P. be d .

Then 2,2 d,2 2d,2 3d,... is the A.P.

10 10d is the sum of the first five terms and 10 35d is the sum of the next five

terms.

According to the conditions given in the question,

1

10 10d 10 35d4

40 40d 10 35d

30 5d

d 6

The th20 term 20a a 20 1 d .

Substitute the values of a and d in the equation to obtain 20a .

20a 2 20 1 6

2 114

112

Therefore, 112 is the th20 term of the A.P.

4. How many terms if the A.P. 11

6, , 5,...2

are needed to give the sum 25 ?

Ans: 25 is the sum of n terms of the given A.P.

The common difference of the A.P. is 11 11 12 1

d 62 2 2


n

nS 2a n 1 d

2 .

Substitute nS 25 , a 6 and

1d

2 in the equation.



n 1

25 2 6 n 12 2

n 150 n 12

2 2

25 n50 n

2 2

100 n 25 n 2n 25n 100 0

Factorize the equation. 2n 5n 20n 100 0

n n 5 20 n 5 0

n 20 or 5

Therefore, 5or 20 terms of the A.P. are needed to give the sum 25 .

5. In an A.P., if th

p term is 1 q and th

q term is 1 p , prove that the sum of first

pq terms is 1

pq 12

, where p q .

Ans:The general term of an A.P. is given by the equation na a n 1 d .


thp term can be written as p

1a a p 1 d

q

and thq term can be written as q

1a a q 1 d

p

Subtract qa from pa .

1 1

p 1 d q 1 dq p

p q

p 1 q 1 dpq

p q

p q dpq

1d

pq



Substitute 1

dpq

in 1

a p 1 dq

.

1 1

a p 1pq q

1 1 1 1a

q q pq pq


n

nS 2a n 1 d

2 .


pq

pq 2 1S pq 1

2 pq pq

1

1 pq 12

1 1pq 1

2 2

1 1pq

2 2

1

pq 12

Therefore, 1

pq 12

is the sum of first pq terms of the A.P.

6. If the sum of a certain number of terms of the A.P. 25,22,19,... is 116 . Find

the last term.

Ans:116 is the sum of n terms of the given A.P.

The common difference of the A.P. is d 22 25 3


n

nS 2a n 1 d

2 .

Substitute nS 116 , a 25 and d 3 in the equation.

n

116 2 25 n 1 32

232 n 50 3n 3



232 n 53 3n 23n 53n 232 0

Factorize the equation. 23n 24n 29n 232 0

3n n 8 29 n 8 0

n 8 3n 29 0

n 8 or 29

3

n 8 as n cannot be equal to 29

3 .

As n 8 the last term is

8a a 8 1 d


8a 25 7 3

25 21

4 Therefore, 4 is the last term of the A.P.

7. Find the sum to n terms of the A.P., whose th

k term is 5k 1 .

Ans:5k 1 is given as the thk term of the A.P.

The equation for thk term of an A.P. is given as ka k 1 d .

Then,

a k 1 d 5k 1

a kd d 5k 1 By comparing the coefficient of k we get the value of d as 5 .

a d 1

a 5 1

a 6 The sum of first n terms of an arithmetic progression is given by the equation

n

nS 2a n 1 d

2 .

Substitute the values of a and d in the equation.

n

nS 2 6 n 1 5

2



n

12 5n 52

n

5n 72

Therefore, n

5n 72

is the sum of n terms of the A.P.

8. If the sum of n terms of an A.P. is 2pn qn , where p and q are constants,

find the common difference.

Ans:The sum of first n terms of an arithmetic progression is given by the equation

n

nS 2a n 1 d

2 .


2n2a n 1 d pn qn

2

2n2a nd d pn qn

2

2 2d dna n n pn qn

2 2

By comparing the coefficients of 2n on both sides, we get the value of d .

That is, d

q2

d 2q

Therefore, 2q is the common difference of the A.P.

9. The sums of n terms of two arithmetic progressions are in the ratio

5n 4:9n 6 . Find the ratio of their th

18 terms.

Ans:Let 1a and

1d be the first term and the common difference of the first arithmetic

progression respectively and 2a and

2d be the first term and the common difference

of the second arithmetic progression respectively.


Sum of n terms of first A.P / Sum of n terms of second A.P. 5n 4

9n 6



1 1

2 2

n2a n 1 d

5n 42n 9n 6

2a n 1 d2

1 1

2 2

2a n 1 d 5n 4

2a n 1 d 9n 6


1 1

2 2

5 35 42a 34d

2a 34d 9 35 6

1 1

2 2

a 17d 179

a 17d 321

We also know that

th18 term of first A.P. / th18 term of second A.P. 1 1

2 2

a 17d

a 17d

Then,

th18 term of first A.P. / th18 term of second A.P. 179

321

Therefore, 179:321 is the ratio of th18 term of both the arithmetic progressions.

10. If the sum of the first p terms of an A.P. is equal to the sum of the first q

terms, then find the sum of the first p q terms.

Ans:Let a be the first term and d be the common difference of the A.P.

The sum of the first p terms of the A.P. is given by the equation

p

pS 2a p 1 d

2 and the first q terms by q

qS 2a q 1 d

2 .


p q

2a p 1 d 2a q 1 d2 2

p 2a p 1 d q 2a q 1 d

2ap pd p 1 2aq qd q 1

2ap p q d p p 1 q q 1 d 0

2 22ap p q d p p q q 0



2ap p q d p q p q p q 0

2ap p q d p q p q 1 0

2a d p q 1 0

2ad

p q 1

Therefore, sum of first p q terms of the A.P. is given by the equation

p q

p qS 2a p q 1 d

2

.

Substituting the value of d in the equation we get,

p q

p q 2aS 2a p q 1

2 p q 1

p q

2a 2a2

0

Therefore, 0 is the sum of the first p q terms of the A.P.

11. Sum of the first p , q and r terms of an A.P. are a , b and c , respectively.

Prove that a b c

q r r p p q 0p q r

.

Ans:Let 1a be the first term and d be the common difference of the A.P.


p 1

pS 2a p 1 d a

2

1

2a2a p 1 d

p

q 1

qS 2a q 1 d b

2

1

2b2a q 1 d

q

r 1

rS 2a r 1 d c

2

1

2c2a r 1 d

r



Subtract qS from pS .

2a 2b

p 1 d q 1 dp q

2aq 2bp

d p 1 q 1pq

2aq 2bp

d p qpq

2 aq bpd

pq p q

Subtract qS from pS .

2b 2c

q 1 d r 1 dq r

2b 2c

d q 1 r 1q r

2br 2qc

d q rqr

2 br qcd

qr q r

Equate both the values of d .

aq bp br qc

pq p q qr q r

qr q r aq bq pq p q br qc

r aq bq q r p br qc p q

aqr bqr q r brp qcp p q

Divide both the sides of the equation by pqr .

a b b c

q r p qp q q r

a b c

q r q r p q p q 0p q r

a b c

q r r p p q 0p q r



Therefore, a b c

q r r p p q 0p q r

is proved.

12. The ratio of the sums of m and n terms of an A.P. is 2 2m :n . Show that the

ratio of thm and th

n term is 2m 1 : 2n 1 .

Ans:Let a be the first term of the A.P. and d be the common difference.


Sum of m terms / Sum of n terms 2

2

m

n

2

2

m2a m 1 d

m2n n

2a n 1 d2

2a m 1 d m

2a n 1 d n

Substitute m 2m 1 and n 2n 1 in the equation.

2a 2m 2 d 2m 1

2a 2n 2 d 2n 1

a m 1 d 2m 1

a n 1 d 2n 1

We also know that

thm term of A.P. / thn term of A.P.

a m 1 d

a n 1 d

Then,

thm term of A.P. / thn term of A.P. 2m 1

2n 1

Therefore, the ratio of thm and thn term is 2m 1 : 2n 1 .

13. If the sum of n terms of an A.P. is 2

3n 5n and its thm term is 164 , find the

value of m .

Ans:Let a be the first term of the A.P. and d be the common difference.

ma a m 1 d 164




n

nS 2a n 1 d

2 .


2n2a n 1 d 3n 5n

2

2n2a nd d 3n 5n

2

2 2d dna n n 3n 5n

2 2

By comparing the coefficients of 2n on both sides, we get the value of d .

That is, d

32

d 6 By comparing the coefficient of n on both sides we get the value of a .

That is, d

a 52

a 3 5

a 8

Substitute the values of a and d in the equation for ma .

8 m 1 6 164

m 1 6 156

m 1 26

m 27 Therefore, 27 is the value of m .

14. Insert five numbers between 8 and 26 such that resulting sequence is an A.P.

Ans:Let the five numbers between 8 and 26 be 1 2 3 4A ,A ,A ,A and

5A . Then the

resulting sequence 1 2 3 4 58,A ,A ,A ,A ,A ,26 is an A.P.

The first term of the A.P. a 8 , the last term b 26 and the number of terms n 7 .

Substitute the values of a,b and n in b a n 1 d .

26 8 7 1 d

6d 26 8

d 3 Then substituting the values of a and d we get



1A a d 8 3 11

2A a 2d 8 2 3 14

3A a 3d 8 3 3 17

4A a 4d 8 4 3 20

5A a 5d 8 5 3 23

Therefore, 11,14,17,20 and 23 are the five numbers between 8 and 26 .

15. If n n

n 1 n 1

a b

a b

is the A.M. between a and b , then find the value of n .

Ans: We know that the A.M. between a and b is given by a b

2

.

According to the conditions given in the question, n n

n 1 n 1

a b a b

2 a b

n 1 n 1 n na b a b 2 a b

n n 1 n 1 n n na ab ba b 2a 2b n 1 n 1 n nab ba a b n 1 n n n 1ab b a a b

n 1 n 1b a b a a b n 1 n 1b a

n 1 0a a

1b b

n 1 0

n 1 Therefore, the value of n is 1 .

16. Between 1 and 31, m numbers have been inserted in such a way that the

resulting sequence is an A.P. and the ratio of th

7 and th

m 1 numbers is 5:9

. Find the value of m .

Ans:Let the m numbers between 1 and 31 be 1 2 mA ,A ,...,A . Then the resulting

sequence 1 2 m1,A ,A ,...,A ,31 is an A.P.

The first term of the A.P. a 1 , the last term b 31 and the number of terms

n m 2 .



Substitute the values of a,b and n in b a n 1 d .

31 1 m 2 1 d

30 m 1 d

30d

m 1

We know that

1A a d

2A a 2d

3A a 3d

Then, the th7 and th

m 1 term is given by the equation,

7A a 7d

m 1A a m 1 d


a 7d 5

a m 1 d 9

301 7

5m 1

30 91 m 1

m 1

m 1 7 30 5

m 1 30 m 1 9

m 1 210 5

m 1 30m 30 9

m 211 5

31m 29 9

9m 1899 155m 145

155m 9m 1899 145

146m 2044

m 14 Therefore, 14 is the value of m .



17. A man starts repaying a loan as first installment of Rs.100 . If he increases

the installment by Rs.5 every month, what amount he will pay in the th30

installment?

Ans:Rs.100 is the first installment of the load and Rs.105 is the second installment

and so on.

Therefore, an A.P. is formed by the amount that the man repays every month.

100,105,110... is the A.P.

The first term of the A.P. a 100 and the common difference d 5 .

The th30 is given by the equation 30A a 30 1 d .

30A 100 29 5

100 145

245

Therefore, Rs. 245 is the amount to be paid in the th30 installment.

18.The difference between any two consecutive interior angles of a polygon is 5

. If the smallest angle is 120 , find the number of the sides of the polygon.

Ans:An A.P. is formed by the angles of the polygon. The first term of the A.P.

a 120 and the common difference d 5 .

We know that 180 n 2 is the sum of all angles of a polygon with n sides.

Therefore, nS 180 n 2

n

2a n 1 d 180 n 22

n

240 n 1 5 180 n 22

n 240 n 1 5 360 n 2

2240n 5n 5n 360n 720 25n 125n 720 0

2n 25n 144 0 2n 16n 9n 144 0

n n 16 9 n 16 0

n 9 n 16 0

n 9 or 16

Therefore, the number of the sides of the polygon 9 or 16 .



Exercise 9.3

1. Find the th20 and th

n term of the G.P. 5 5 5

, , , ...2 4 8

Ans:5 5 5

, , ,...2 4 8

is the given G.P.

The first term of the G.P. is 5

a2

and the common ratio is 5 4 1

r5 2 2

.

The thn term of the G.P. is given by the equation n 1

na ar .

Substituting the values of a and r we get

n 1

n n 1 n

5 1 5 5a

2 2 2 2 2

Similarly, the th20 term of the G.P. is 20 1

20a ar

19

20 19 20

5 1 5 5a

2 2 2 2 2

Therefore, the th20 and thn term of the given G.P. is

20

5

2 and

n

5

2 respectively.

2. Find the th

12 term of a G.P. whose th

8 term is 192 and the common ratio is 2

Ans:Let the first term of the G.P. be a and the common ratio r 2 .

The th8 term of the G.P. is given by the equation 8 1

8a ar .

Substituting the values of 8a and r we get

7

192 a 2

6 7

2 3 a 2

6

7

2 3 3a

22

Then th12 term of the G.P. is given by the equation 12 1

12a ar .

Substitute the values of a and r in the equation.

11

12

3a 2

2

10

3 2



3072

Therefore, the th12 term of the G.P. is 3072 .

3. The th5 , th

8 and th11 terms of a G.P. are p ,q and s , respectively. Show that

2q ps .

Ans:Let the first term and the common ratio of the G.P. be a and r respectively.

According to the conditions given in the question, 5 1 4

5a ar ar p 8 1 7

8a ar ar q 11 1 10

11a ar ar s

Dividing 8a by

5a we get 7

4

ar q

ar p

3 qr

p

Dividing 11a by

8a we get 10

7

ar s

ar q

3 sr

q

Equate both the values of 3r obtained.

q s

p q

2q ps

Therefore, 2q ps is proved.

4. The th

4 term of a G.P. is square of its second term, and the first term is 3 .

Determine its th

7 term.

Ans:Let the first term and the common ratio of the G.P. be a and r respectively.

It is given that a 3 .


na ar .

Then,



3 3

4a ar 3 r

1

2a ar 3 r


233 r 3 r

3 23r 9r

7r 3a 6ar

6

3 3

7

3

2187 Therefore, 2187 is the seventh term of the G.P.

5. Which term of the following sequences:

a) 2,2 2,4... is 128 ?

Ans:2,2 2,4... is the given sequence.

The first term of the G.P. a 2 and the common ratio r 2 2 2 2 .

128 is the thn term of the given sequence.


na ar .

Therefore, n 1ar 128

n 1

2 2 128

n 1

722 2 2

n 1

1 722 2

n 11 7

2

n 16

2

n 1 12

n 13

Therefore, 128 is the th13 term of the given sequence.



b) 3,3,3 3... is 729 ?

Ans: 3,3,3 3... is the given sequence.

The first term of the G.P. a 3 and the common ratio r 3 3 3 .



na ar .

Therefore, n 1ar 729

n 1

3 3 729

n 1

1 2 623 2 3

1 n 1

62 23 3

1 n 16

2 2

1 n 16

2

n6

2

n 12

Therefore, 729 is the th12 term of the given sequence.

c) 1 1 1

, , , ...3 9 27

is 1

19683 ?

Ans:1 1 1

, , ,...3 9 27

is the given sequence.

The first term of the G.P. 1

a3

and the common ratio 1 1 1

r9 3 3

.

1



na ar .

Therefore, n 1 1ar

19683

n 11 1 1

3 3 19683



n 91 1

3 3

n 9

Therefore, 1

19683 is the th9 term of the given sequence.

6. For what values of x , the numbers 2 7

,x,7 2

are in G.P.?

Ans:2 7

,x,7 2

are the given numbers and the common ratio x 7x

2 7 2

We also know that, common ratio 7 2 7

x 2x

Equating both the common ratios we get

7x 7

2 2x

2 2 7x 1

2 7

x 1

x 1 Therefore, the given numbers will be in G.P. for x 1 .

7. Find the sum up to 20 terms in the geometric progression 0.15,0.015,0.0015...

Ans:0.15,0.015,0.0015... is the given G.P.

The first term of the G.P. a 0.15 and the common ratio 0.015

r 0.10.15

.

The sum of first n terms of the G.P. is given by the equation n

n

a 1 rS

1 r

.

Therefore, the sum of first 20 terms of the given G.P. is

20

20

0.15 1 0.1S

1 0.1

200.15

1 0.10.9



2015

1 0.190

201

1 0.16

Therefore, the sum up to 20 terms in the geometric progression 0.15,0.015,0.0015...

is 201

1 0.16

.

8. Find the sum of n terms in the geometric progression 7, 21,3 7...

Ans: 7, 21,3 7... is the given G.P.

The first term of the G.P. a 7 and the common ratio 21

r 37

.

The sum of the first n terms of the G.P. is given by the equation n

n

a 1 rS

1 r

.

The sum of the first n terms of the given G.P. is

n

n

7 1 3

S1 3

n

7 1 31 3

1 3 1 3

n

7 3 1 1 3

1 3

n

7 3 1 1 3

2

n

2

7 1 33 1

2

Therefore, the sum of n terms of the geometric progression 7, 21,3 7... is

n

2

7 1 33 1

2

.



9. Find the sum of n terms in the geometric progression 2 3

1, a,a , a ... ( if

a 1 )

Ans:2 31, a,a , a ... is the given G.P.

The first term of the G.P. 1a 1 and the common ratio r a .


1

n

a 1 rS

1 r

.

The sum of first n terms of the given G.P. is

n

n

1 1 aS

1 a

n

1 a

1 a

Therefore, the sum of n terms of the geometric progression 2 31, a,a , a ... is

n

1 a

1 a

.

10. Find the sum of n terms in the geometric progression 3 5 7

x ,x ,x ...( if a 1 )

Ans:3 5 7x ,x ,x ... is the given G.P.

The first term of the G.P. 3a x and the common ratio 2r x .


1

n

a 1 rS

1 r

.

The sum of first n terms of the given G.P. is

n3 2

n 2

x 1 xS

1 x

3 2n

2

x 1 x

1 x

Therefore, the sum of n terms of the geometric progression 3 5 7x ,x ,x ... is

3 2n

2

x 1 x

1 x

.



11. Evaluate 11

k 1

2 3k

Ans: 11 11 11 11

k

k 1 k 1 k 1 k 1

2 3k (2) (3k) 22 3

…(1)

We know that, 11

k 1 2 11

k 1

(3 ) 3 3 3

This sequence 2 3 113,3 ,3 , ,3 forms a G.P. Therefore,

n

n

a r 1S

r 1

Substituting the values to the above equation we get,

11

n

3 3 1S

3 1

11

n

3S 3 1

2

Therefore,

11

k 11

k 1

33 3 1

2

Substitute this value in equation (1).

11

11

k 1

32 3k 22 3 1

2

12. The sum of first three terms of a G.P. is 39

10 and their product is 1 . Find the

common ratio and the terms.

Ans:Let the first three terms of a G.P. be a

,a,arr

.

Then, its sum is

a 39a ar

r 10 …(1)

And the product is



a

a ar 1r

…(2)

Solving equation (2) we will get, 3a 1

Considering the real roots,

a 1 Substitute the value of a in the equation.

1 391 r

r 10

2 391 r r r

10

210 10r 10r 39r 210r 29r 10 0 210r 25r 4r 10 0

5r 2r 5 2 2r 5 0

5r 2 2r 5 0

2r

5 or

5

2

Therefore, 5

,12

and 2

5 are the first three terms of the G.P.

13. How many terms of G.P. 2 3

3,3 ,3 ... are needed to give the sum 120?

Ans:Given G.P. 2 3 113,3 ,3 , ,3 .

Let there be n terms to get the sum as 120 .

Then using the formula, we get,

n

n

a r 1S

r 1

…(1)

Given that,

nS 120

a 3

r 3

Substituting the given values in equation (1),



n

n

3 3 1S 120

3 1

n3 3 1120

2

n120 23 1

3

n3 1 80 n3 81 n 43 3

n 4 Therefore, for getting the sum as 120 the given G.P. should have 4 terms.

14. The sum of first three terms of a G.P. is 16 and the sum of the next three

terms is 128 . Determine the first term, the common ratio, and the sum to n terms

of the G.P.

Ans:Let 2 3a,ar,ar ,ar ... be the G.P.

According to the conditions given in the question, 2a ar ar 16 …(1)

3 4 5ar ar ar 128 …(2)

Equation (1) and (2) can also be written as,

2a 1 r r 16

3 2ar 1 r r 128

Divide equation (2) by (1) .

3 2

2

ar 1 r r2 128

1 16a 1 r r

3r 8

r 2 Substituting the value of r in equation (1), we get

2a 1 r r 16

a 1 2 4 16

7a 16



16a

7

Sum of n terms of the G.P. is,

n

n

a r 1S

r 1

n

n

2 116S

7 2 1

n

n

16S 2 1

7

Therefore, the first term of the G.P. is 16

a7

, the common ratio r 2 and the sum of

terms n

n

16S 2 1

7 .

15. Given a G.P. with a 729 and th

7 term 64 , determine 7

S .

Ans:Given thata 729 and 7a 64

Let the common ratio of the G.P be r . Then, n 1

na ar 6 1

7a ar

664 729 r

6

6 2r

3

2r

3

We know that,

n

n

a 1 rS

1 r

Therefore,



7

7

2729 1

3S

21

3

7 7

7

3 2729 3

3

7 7

7

7

3 23

3

7 7

3 2

2187 128

2059

Therefore, the value of 7S is 2059 .

16. Find a G.P. for which sum of the first two terms is 4 and the fifth term is 4

times the third term.

Ans:Let a and r be the first term and common ratio of the G.P. respectively.


5 3a 4 a 4 2ar 4 ar

2r 4

r 2 Given that,

2

2

a 1 rS 4

1 r

Substituting r 2 in the above equation,

2

a 1 24

1 2

a 1 44

1

4 a 3



4a

3

Now, taking r 2 , we get,

2a 1 2

41 2

a 1 44

1 2

a 34

3

a 4

Therefore , 4 8 16

, , ,...3 3 3

or 4, 8, 16, 32... is the required G.P.

17. If the th

4 , th10 and th

16 terms of a G.P. are x,y and z , respectively. Prove

that x,y,z are in G.P.

Ans:Let the first term of the G.P be a and the common ratio be r .

According to the conditions given in the question, 3

4a ar x …(1) 9

10a ar y …(2) 15

16a ar z …(3)

Then divide equation (2) by (1) . 9

3

y ar

x ar

6yr

x

Now, divide equation (3) by (1). 15

9

z ar

y ar

6zr

y

Therefore,

y z

x y



Therefore, it is proved that x,y,z are in G. P.

18. Find the sum to n terms of the sequence, 8,88,888,8888...

Ans:8,88,888,8888... is the given sequence

The given sequence is not in G.P. In order to make the sequence in G.P., it has to be

changed to the form,

nS 8 88 888 8888 ... to n terms

8

9 [9 99 999 9999 ... to n terms]

8

9 [ 2 3 410 1 10 1 10 1 10 1 to n terms]

8

9 [( 210 10 ...n terms) ( 1 1 1 ...n terms)]

n10 10 18n

9 10 1

n10 10 18n

9 9

n80 810 1 n

81 9

Therefore, the sum of n terms the given sequence is n80 810 1 n

81 9 .

19. Find the sum of the products of the corresponding terms of the sequences

2,4,8,16,32 and 128,32,8,2,1 2 .

Ans:1

2 128 4 32 8 8 16 2 322

2

1 164 4 2 1

2 2

is the required sum.

We can see that, 2

1 14,2,1, ,

2 2 is a G.P.

The first term of the G.P. is a 4 and the common ratio is 1

r2

.



We know that,

n

n

a 1 rS

1 r

Therefore, 5

3

14 1

2S

11

2

14 1

32

1

2

32 18

32

31

4

Therefore, the required sum 31

64 16 31 4964

.

20. Show that the products of the corresponding terms of the sequences form 2 n 1

a,ar,ar ,...ar and

2 n 1A,AR,AR ,AR

a G.P. and find the common ratio.

Ans: The sequence 2 2 n 1 n 1aA,arAR,ar AR ,...ar AR forms a G.P. is to be proved.

Second term / First term arAR

rRaA

Third term / Second term 2 2ar AR

rRaA

Therefore, the 2 2 n 1 n 1aA,arAR,ar AR ,...ar AR

forms a G.P. and the common ratio is

rR .

21. Find four numbers forming a geometric progression in which third term is

greater than the first term by 9 , and the second term is greater than the th

4 by

18 .

Ans:Let the first term be a and the common ratio be r of the G.P.



2 3

1 2 3 4a a,a ar,a ar ,a ar


3 1a a 9 2ar a 9

2a r 1 9 ...(1)

2 4a a 9

3ar ar 18

2ar 1 r 18 ...(2)

Divide (2) by (1).

2

2

ar 1 r 18

9a r 1

r 2

r 2 Substitute r 2 in equation (1).

a 4 1 9

a 3 9

a 3

Therefore, 2

3,3 2 ,3 2 and 3

3 2 ,i.e., 3, 6,12 and 24 are the first four

numbers of the G.P.

22. If th

p , th

q and th

r terms of a G.P. are a,b and c , respectively. Prove that q r r p p q

a b c 1 .

Ans:Let the first term be A and the common ration be R of the G.P.

According to the conditions given in the question, p 1AR a q 1AR b r 1AR c

Then, q r r p p qa b c

p 1 q r q 1 r p r 1 p qq r r p q rA R A R A R

pq pr q r rq r p pq pr p qr qq r r p p qA R

0 0A R

1



Therefore, q r r p p qa b c 1 is proved.

23. If the first and thn the term of a G.P. are a and b , respectively, and if P is

the product of n terms, prove that n2

P ab .

Ans:a is the first term and b is the last term of the G.P.

Therefore, is the G.P. 2 3 n 1a,ar,ar ,ar ...ar , where the common ratio is r . n 1b ar …(1)

P is the product of n terms. Therefore,

2 n 1P a ar ar ... ar

2 n 1a a ...a r r ...r

1 2 ... n 1na r

…(2)

We can see that, 1,2,... n 1 is an A.P. Therefore,

1 2 ... n 1

n 1

2 n 1 1 12

n 1

2 n 22

n n 1

2

So, equation (2) can be written as n n 1

n 2P a r

.

Therefore, n n 12 2nP a r

n

n 12a r

nn 1a ar

Substituting (1) in the equation,

n2P ab

Therefore, n2P ab is proved.



24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms

from th

n 1 to th

2n term is n

1

r .

Ans:Let the first term be a and the common ration be r of the G.P.

na 1 r

1 r

is the sum of first n terms.

From th

n 1 to th

2n term there are n terms.

From th

n 1 to th

2n term the sum of the terms is

n

n 1

n

a 1 rS

1 r

n 1 n 1 1 na ar ar

Therefore, the required ratio is

n

nn n

a 1 r 1 r 1

1 r rar 1 r

Therefore, n

1

r is the ratio of the sum of first n terms of a G.P. to the sum of terms

from th

n 1 to th

2n term .

25. If a,b,c and d are in G.P. show that:

22 2 2 2 2 2

a b c b c d ab bc cd

Ans:Let us assume a,b,c,d are in G.P.

Therefore,

bc ad …(1) 2b ac …(2) 2c bd …(3)

To prove :

22 2 2 2 2 2a b c b c d ab bc cd

R.H.S.

2

ab bc cd

Substitute (1) in the equation.

2

ab ad cd



2

ab d a c

22 2 2a b 2abd a c d a c

2 2 2 2 2 2a b 2a bd 2acbd d a 2ac c

Substitute (1) and (2) in the equation. 2 2 2 2 2 2 2 2 2 2 2 2 2 2a b 2a c 2b c d a d b d b d c 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2a b a c a c b c b c d a d b d b d c 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2a b a c a d b b b c b d c b c c c d

Substitute (2) and (3) in the equation and rearrange the terms.

2 2 2 2 2 2 2 2 2 2 2 2a b c d b b c d c b c d

2 2 2 2 2 2a b c b c d

L.H.S. Therefore, L.H.S. R.H.S.

Therefore, 22 2 2 2 2 2a b c b c d ab bc cd is proved.

26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Ans:Let the two numbers between 3 and 81 be 1G and

2G such that the series,

1 23,G ,G ,81 , forms a G.P.

Let the first term be a and the common ration be r of the G.P.

Therefore,

3

81 3 r 3r 27

Taking the real roots, we get r 3 .

When r 3 ,

1G ar 3 3 9

22

2G ar 3 3 27

Therefore, 9 and 27 are the two required numbers.

27. Find the value of n so that n 1 n 1

n n

a b

a b

may be the geometric mean between

a and b .

Ans:The geometric mean of a and b is ab .

According to conditions given in the question,



n 1 n 1

n n

a bab

a b

Square on both the sides.

2n 1 n 1

2n n

a bab

a b

2n 2 n 1 n 1 2n 2 2n n n 2na 2a b b ab a 2a b b

2n 2 n 1 n 1 2n 2 2n 1 n 1 n 1 2n 1a 2a b b a b 2a b ab 2n 2 2n 2 2n 1 2n 1a b a b ab 2n 2 2n 1 2n 1 2n 2a a b ab b

2n 1 2n 1a a b b a b 2n 1 0

a a1

b b

2n 1 0

1n

2

28. The sum of two numbers is 6 times their geometric mean, show that numbers

are in the ratio 3 2 2 : 3 2 2 .

Ans:Let a and b be the two numbers.

ab is the geometric mean.


a b 6 ab …(1)

2

a b 36ab

Also,

2 2

a b a b 4ab 36ab 4ab 32ab

a b 32 ab

4 2 ab …(2)

Add (1) and (2).

2a 6 4 2 ab

a 3 2 2 ab



Substitute a 3 2 2 ab in equation (1) .

b 6 ab 3 2 2 ab

b 3 2 2 ab

Divide a by b .

3 2 2 aba 3 2 2

b 3 2 23 2 2 ab

Therefore, it is proved that the numbers are in the ratio 3 2 2 : 3 2 2 .

29. If A and B be A.M. and G.M., respectively between two positive numbers,

prove that the numbers are A A G A G .

Ans:Given: The two positive numbers between A.M. and G.M. are A and G .

Let a and b be these two positive numbers.

Therefore, a b

AM A2

…(1)

GM G ab …(2)

Simplifying (1) and (2) , we get

a b 2A …(3) 2ab G …(4)

Substituting (3) and (4) in the identity,

2 2

a b a b 4ab ,

We get

2 2 2 2 2a b 4A 4G 4 A G

2

a b 4 A G A G

a b 2 A G A G …(5)

Adding (3) and (5) we get ,

2a 2A 2 A G A G

a A A G A G

Substitute a A A G A G in equation (3).



b 2A A A G A G

A A G A G

Therefore, A A G A G are the two numbers.

30. The number of bacteria in a certain culture doubles every hour. If there were

30 bacteria present in the culture originally, how many bacteria will be present

at the end of nd2 hour, th

4 hour and thn hour?

Ans:The number of bacteria after every hour will form a G.P. as it is given that the

number of bacteria doubles every hour.

Given: a 30 and r 2

Therefore,

22

3a ar 30 2 120

That is, 120 will be the number of bacteria at the end of nd2 hour.

44

5a ar 30 2 480

That is, 480 will be the number of bacteria at the end of th4 hour.

n n

n 1a ar 30 2

Therefore, n

30 2 will be the number of bacteria at the end of thn hour.

31. What will Rs.500 amounts to in 10 years after its deposit in a bank which

pays annual interest rate of 10% compounded annually?

Ans:Rs.500 is the amount deposited in the bank.

The amount Rs.1

500 110

Rs. 500 1.1 , at the end of first year.

The amount Rs. 500 1.1 1.1 , at the end of nd2 year.

The amount Rs. 500 1.1 1.1 1.1 , at the end of rd3 year and so on.

Therefore, the amount at the end of 10 years

Rs. 500 1.1 1.1 ...(10 times)

Rs. 10

500 1.1

32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5 , respectively,

then obtain the quadratic equation.

Ans:Let a and b be the root of the quadratic equation.




a bA.M. 8

2

a b 16 …(1)

G.M. ab 5

ab 25 …(2)

The quadratic equation is given by the equation, 2x x (Sum of roots) (Product of roots) 0

2x x a b ab 0

Substituting (1) and (2) in the equation. 2x 16x 25 0

Therefore, 2x 16x 25 0 is the required quadratic equation.

Exercise 9.4

1. Find the sum to n terms of the series 1 2 2 3 3 4 4 5 ...

Ans:1 2 2 3 3 4 4 5 ... is the given series.

The thn term of the series is na n n 1 .

The sum of n terms of a series is given by the equation n

n k

k 1

S a

.

The sum of n terms of the given series is

n

n

k 1

S k k 1

n n2

k 1 k 1

k k

n n 1 2n 1 n n 1

6 2

n n 1 2n 11

2 3

n n 1 2n 4

2 3

n n 1 n 2

3



Therefore, the sum of n terms of the series 1 2 2 3 3 4 4 5 ... is

n n 1 n 2

3

.

2. Find the sum to n terms of the series 1 2 3 2 3 4 3 4 5 ...

Ans:1 2 3 2 3 4 3 4 5 ... is the given series.

The thn term of the series is

na n n 1 n 2

2n n n 2

3 2n 3n 2n


n k

k 1

S a

.


n

3 2

n

k 1

S k 3k 2k

n n n3 2

k 1 k 1 k 1

k 3 k 2 k

2

n n 1 3n n 1 2n 1 2n n 1

2 6 2

2

n n 1 n n 1 2n 1n n 1

2 2

n n 1 n n 12n 1 2

2 2

2n n 1 n n 4n 6

2 2

2n n 1n 5n 6

4

2n n 1n 2n 3n 6

4

n n 1 n n 2 3 n 2

4



n n 1 n 2 n 3

4

Therefore, the sum of n terms of the series 1 2 3 2 3 4 3 4 5 ... is

n n 1 n 2 n 3

4

.

3. Find the sum to n terms of the series 2 2 23 1 5 2 7 3 ...

Ans: 2 2 23 1 5 2 7 3 ... is the given series.


2

na 2n 1 n 3 22n n


n k

k 1

S a

.


n

3 2

n

k 1

S 2k k

n n3 2

k 1 k 1

2 k k

2

n n 1 n n 1 2n 12

2 6

22n n 1 n n 1 2n 1

2 6

n n 1 2n 1n n 1

2 3

2n n 1 3n 3n 2n 1

2 3

2n n 1 3n 5n 1

2 3

2n n 1 3n 5n 1

6



Therefore, the sum of n terms of the series 2 2 23 1 5 2 7 3 ... is

2n n 1 3n 5n 1

6

.

4. Find the sum to n terms of the series 1 1 1

...1 2 2 3 3 4

Ans:1 1 1

...1 2 2 3 3 4

is the given series.


n

1a

n n 1

By partial fractions the above equation can be written as

n

1 1a

n n 1

Therefore,

1

1 1a

1 2

2

1 1a

2 3

3

1 1a ...

3 4

n

1 1a

n n 1

Add the above terms.

1 2 n

1 1 1 1 1 1 1 1a a ... a ... ...

1 2 3 n 2 3 4 n 1


n k

k 1

S a

.


n

1S 1

n 1

n 1 1

n 1



n

n 1

Therefore, the sum of n terms of the series 1 1 1

...1 2 2 3 3 4

is n

n 1 .

5. Find the sum to n terms of the series 2 2 2 25 6 7 ... 20

Ans: 2 2 2 25 6 7 ... 20 is the given series.


2

na n 4 2n 8n 16


n k

k 1

S a

.


n

2

n

k 1

S k 8k 16

n n2

k 1 k 1

k 8 k 16

n n 1 2n 1 8n n 116n

6 2

Then,

2220 16 4 is the th16 term.

Therefore,

16

16 16 1 2 16 1 8 16 16 1S 16 16

6 2

16 17 33 8 16 17256

6 2

1496 1088 256

2840

Therefore, the sum of n terms of the series 2 2 2 25 6 7 ... 20 is 2840 .

6. Find the sum to n terms of the series 3 8 6 11 9 14 ...

Ans:n3 8 6 11 9 14 ...a is the given series.




na ( thn term of 3,6,9...) ( thn term of 8,11,14...)

3n 3n 5 29n 15n


n k

k 1

S a

.


n

2

n

k 1

S 9k 15k

n n2

k 1 k 1

9 k 15 k

n n 1 2n 1 n n 19 15

6 2

3n n 1 2n 1 15n n 1

2 2

3n n 12n 1 5

2

3n n 12n 6

2

3n n 1 n 3

Therefore, the sum of n terms of the series 3 8 6 11 9 14 ... is

3n n 1 n 3 .

7. Find the sum to n terms of the series 2 2 2 2 2 21 1 2 1 2 3 ...

Ans: 2 2 2 2 2 2

n1 1 2 1 2 3 ...a is the given series.


2 2 2 2

na 1 2 3 ... n

n n 1 n 1

6

2n 2n 3n 1

6



3 22n 3n n

6

3 21 1 1n n n

3 2 6


n k

k 1

S a

.


n

3 2

n

k 1

1 1 1S k k k

3 2 6

n n n3 2

k 1 k 1 k 1

1 1 1k k k

3 2 6

22

2

n n 1 n n 1 2n 1 n n 11 1 1

3 2 6 6 22

n n 1 n n 1 2n 1 1

6 2 2 2

2n n 1 n n 2n 1 1

6 2

2n n 1 n n 2n 2

6 2

n n 1 n 1 2 n 1

6 2

n n 1 n 1 n 2

6 2

2

n n 1 n 2

12

Therefore, the sum of n terms of the series 2 2 2 2 2 21 1 2 1 2 3 ... is

2

n n 1 n 2

12

.



8. Find the sum to n terms of the series whose thn term is given by n n 1 n 4

Ans:The thn term of the series is

na n n 1 n 4

2n n 5n 4

3 2n 5n 4n


n k

k 1

S a

.


n n n

3 2

n

k 1 k 1 k 1

S k 5 k 4 k

22n n 1 5n n 1 2n 1 4n n 1

4 6 2

n n 1 n n 1 5 2n 14

2 2 3

2n n 1 3n 3n 20n 10 24

2 6

2n n 1 3n 23n 34

2 6

2n n 1 3n 23n 34

12

Therefore, the sum of thn terms of the series whose thn term is given by

n n 1 n 4 is 2n n 1 3n 23n 34

12

.

9. Find the sum to n terms of the series whose thn term is given by

2 nn 2

Ans:The thn term of the series is 2 n

na n 2


n k

k 1

S a

.

The sum of first n terms of the given series is



n

2 k

n

k 1

S k 2

n n2 k

k 1 k 1

k 2

Let n

k 1 2 3

k 1

2 2 2 2 ...

Both the first term and the common ratio of 1 2 32 2 2 ... which forms a G.P. is 2 .

Therefore,

n

nk

k 1

2 2 12

2 1

n2 2 1

Then,

n

2 n

n

k 1

S k 2 2 1

nn n 1 2n 1

2 2 16

Therefore, the sum of n terms of the series whose thn term is given by 2 nn 2 is

nn n 1 2n 1

2 2 16

.

10. Find the sum to n terms of the series whose thn term is given by

22n 1

Ans:The thn term of the series is

2

na 2n 1 24n 4n 1


n k

k 1

S a

.

The sum of first n terms of the given series is

n

2

n

k 1

S 4k 4k 1

n n n2

k 1 k 1 k 1

4 k 4 k 1



4n n 1 2n 1 4n n 1n

6 2

2n n 1 2n 12n n 1 n

3

22 2n 3n 1n 2 n 1 1

3

24n 6n 2 6n 6 3n

3

24n 1n

3

n 2n 1 2n 1

3

Therefore, the sum of n terms of the series whose thn term is given by 2

2n 1 is

n 2n 1 2n 1

3

.

Miscellaneous Exercise

1. Show that the sum of th

m n and th

m n terms of an A.P. is equal to twice

the th

m term.

Ans:Let the first term of the A.P. be a and the common difference be d .

The term of an A.P. is given by the equation

ka a k 1 d

Therefore,

m na a m n 1 d

m na a m n 1 d

ma a m 1 d

Add m na

and m na

.

m n m na a a m n 1 d a m n 1 d

2a m n 1 m n 1 d

2a 2m 2 d



2a 2 m 1 d

2 a m 1 d

m2a

Therefore, the sum of th

m n and th

m n terms of an A.P. is equal to twice the thm term is proved.

2. Let the sum of three numbers in A.P., is 24 and their product is 440, find the

numbers.

Ans:Let a d , a and a d be the three numbers in A.P.


a d a a d 24

3a 24

a 8

a d a a d 440

Substituting a 8 in the equation.

8 d 8 8 d 440

8 d 8 d 55 264 d 55

2d 64 55 2d 9

d 3 When d 3 ,

the three numbers are 5,8 and 11.

When d 3 ,

the three numbers are 11,8 and 5 .

Therefore, 5,8 and 11 are the three numbers.

3. Let the sum of n,2n,3n terms of an A.P. be 1 2

S ,S and 3

S , respectively, show

that 3 2 1S 3 S S .

Ans:Let the first term of the A.P. be a and the common difference be d .


n

nS 2a n 1 d

2 .

Therefore,



1

nS 2a n 1 d

2

2

2nS 2a n 1 d n 2a n 1 d

2

n

3nS 2a n 1 d

2

Subtract 1S from

2S .

2 1 2

nS S S n 2a n 1 d 2a n 1 d

2

4a 4nd 2d 2a nd dn

2

2a 3nd dn

2

n

2a 3n 1 d2

Therefore,

3

3nS 2a 3n 1 d

2

2 13 S S

Therefore, 3 2 1S 3 S S is proved.

4. Find the sum of all numbers between 200 and 400 which are divisible by 7 .

Ans:203,210,217,...,399 are the numbers lying between 200 and 400 , which are

divisible by 7 . An A.P. is formed by this series.

The first term of the A.P. is a 203 , the last term is na 399 and the common

difference is d 7 .




203 n 1 7 399

n 1 7 196

n 1 28

n 29




n n

nS a a

2 .


29

29S 203 399

2

29

6022

29 301

8729 Therefore, 8729 is the sum of numbers lying between 200 and 400 , which are

divisible by 7 .

5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5 .

Ans:2,4,6,...100 are the integers lying between 1 to 100 , which are divisible by 2 .

An A.P. is formed by this series.

The first term of the A.P. and the common difference is equal to 2 .



100 2 n 1 2


n

nS 2a n 1 d

2 .


50

2 4 6 ... 100 2 2 50 1 22

50

4 982

25 102

2550

5,10,...100 are the integers lying between 1 to 100 , which are divisible by 5 . An A.P.

is formed by this series.






100 5 n 1 5

5n 100


n

nS 2a n 1 d

2 .


20

5 10 ... 100 2 5 20 1 52

10 10 19 5

10 10 95

10 105

1050

10,20,...100 are the integers lying between 1 to 100 , which are divisible by 2 and 5

. An A.P. is formed by this series.




100 10 n 1 10

100 10n


n

nS 2a n 1 d

2 .


10

10 20 ... 100 2 10 10 1 102

5 20 90

5 110

550 Therefore,

2250 1050 550 3050 is the required sum.



Therefore, 3050 is the sum of the integers from 1 to 100 , which are divisible by 2 or

5 .

6. Find the sum of all two-digit numbers which when divided by 4 , yields 1 as

remainder.

Ans:13,17,...97 are the two-digit numbers which when divided by 4 yields 1 as

remainder. An A.P. is formed by this series.

The first term of the A.P. is a 13 , the last term is na 97 and the common

difference is d 4 .



97 13 n 1 4

4 n 1 84

n 1 21


n

nS 2a n 1 d

2 .


22

22S 2 13 22 1 4

2

11 26 84

1210 Therefore, 1210 is the sum of all two-digit numbers which when divided by 4 , yields

1 as remainder.

7. If is a function satisfying f x y f x .f y for all x,y N , such that

f 1 3 and n

x 1

f x 120

find the value of n .

Ans:According to the given conditions in the question,

f x y f x f y for all x,y, N

f 1 3

Let x y 1 .

Then,



f 1 1 f 1 2 f 1 f 2 3 3 9

We can also write

f 1 1 1 f 3 f 1 2 f 1 f 2 3 9 27

f 4 f 1 4 f 1 f 3 3 27 81

Both the first term and common ratio of f 1 ,f 2 ,f 3 ,..., that is 3,9,27,..., that

forms s G.P. is equal to 3

We know that, n

n

a r 1S

r 1

Given that, n

k 1

f x 120

Then,

n3 3 1120

3 1

n3120 3 1

2

n3 1 80 n 43 80 3 n3 1 80

n 4 Therefore, 4 is the value of n .

8. The sum of some terms of G.P. is 315 whose first term and the common ratio

are 5 and 2 , respectively. Find the last term and the number of terms. Ans:Let 315 be the sum of n terms of the G.P.

We know that, n

n

a r 1S

r 1

The first term a of the A.P. is 5 and the common difference r is 2 .

Substitute the values of a and r in the equation

n5 2 1315

2 1

n2 1 63

2n2 63 2

n 6



Therefore, the th6 term is the last term of the G.P.

th6 term 56 1ar 5 2 5 32 160

Therefore, 160 is the last term of the G.P and the number of terms is 6 .

9. The first term of a G.P. is 1 . The sum of the third term and fifth term is 90 .

Find the common ratio of G.P.

Ans:Let the first term of the G.P. be a and the common ratio be r .

Then, a 1 2 2

3a ar r 4 4

5a ar r

Therefore, 2 4r r 90

4 2r r 90 0

2 1 1 360r

2

1 361

2

10 or 9

r 3 Therefore, 3 is the common ratio of the G.P.

10. The sum of the three numbers in G.P. is 56 . If we subtract 1,7,21 from

these numbers in that order, we obtain an arithmetic progression. Find the

numbers.

Ans:Let a,ar and 2ar be the three numbers in G.P.

According to the conditions given in the question, 2a ar ar 56

2a 1 r r 56 …(1)

An A.P. is formed by 2a 1,ar 7,ar 21

Therefore,

2ar 7 a 1 ar 21 ar 7 b

2ar a 6 ar ar 14 2ar 2ar a 8



2ar ar ar a 8

2a r 1 2r 8

2

2a r 1 8 …(2)

Equating (1) and (2), we get

2 27 r 2r 1 1 r r

2 27r 14r 7 1 r r 26r 15r 6 0 26r 12r 3r 6 0

6 r 2 3 r 2 0

6r 3 r 2 0

Then,8,16 and 32 are the three numbers when r 2 and 32,16 and 8 are the numbers

when 1

r2

.

Therefore, 8,16 and 32 are the three required numbers in either case.

11.A G.P. consists of an even number of terms. If the sum of all the terms is 5

times the sum of terms occupying odd places, then find its common ratio.

Ans: Let 1 2 3 4 2nT ,T ,T ,T ,...T be the G.P.

2n is the number of terms.


1 2 3 2n 1 3 2n 1T T T ... T 5 T T ... T

1 2 3 2n 1 3 2n 1T T T ... T 5 T T ... T 0

2 4 2n 1 3 2n 1T T ... T 4 T T ... T

Let 2 3a,ar,ar ,ar be the G.P.

Therefore,

n nar r 1 4 a r 1

r 1 r 1

ar 4a

r 4

Therefore, 4 is the common ratio of the G.P.

12. The sum of the first four terms of an A.P. is 56 . The sum of the last four

terms is 112 . If its first term is 11, then find the number of terms.



Ans:Let a,a d,a 2d,a 3d...a n 2 d,a n 2 d be the A.P.

a a d a 2d a 3d 4a 6d is the sum of the first four terms.

a n 4 d a n 3 d a n 2 d a n 1 d 4a 4n 10 d

is the sum of the last four terms.


4a 6d 56 It is given that a 11 , then

4 11 6d 56

6d 12

d 2 Therefore,

4a 4n 10 d 112

4 11 4n 10 2 112

4n 10 2 68

4n 10 34

4n 44

n 11 Therefore, 11 is the number of terms of the A.P.

13: If a bx b cx c dx

x 0a bx b cx c dx

then show that a,b,c and d are in G.P.

Ans:Given ,

a bx b cx

a bx b cx

a bx b cx b cx a bx 2 2 2 2ab acx b x bcx ab b x acx bcx

22b x 2acx 2b ac

b c

a b

It is also given that,

b cx c dx

b cx c dx

b cx c dx b cx c dx



2 2 2 2bc bdx c x cdx bc bdx c x cdx 22c x 2bdx

2c bd

c d

d c

Equating both the results, we get

b c d

a b b

Therefore, it is proved that a,b,c and d are in G.P.

14. Let S be the sum, P the product and R the sum of reciprocals of terms in a

G.P. Prove that 2 n nP R S .

Ans:Let 2 3 n 1a,ar,ar ,ar ...ar be the G.P.


na r 1S

r 1

n 1 2 ... n 1P a r

Since the sum of first n natural numbers is n 1

n2

n n 1

n 2P a r

n 1

1 1 1R ...

a ar ar

n 1 n 2

n 1

r r ...r 1

ar

Since n 11,r,...r

forms a G.P.,

n

n 1

1 r 1 1R

r 1 ar

n

n 1

r 1

ar r 1

Then,



nn

n n 12 n 2n

nn n 1n

r 1P R a r

a r r 1

nn n

n

a r 1

r 1

nna r 1

r 1

nS

Therefore, 2 n nP R S .

15. The th

p ,th

q and th

r terms of an A.P. are a,b,c respectively. Show that

q r a r p b p q c 0 .

Ans:Let the first term of the A.P. be t and the common difference be d .

na t n 1 d is the equation of the thn term of the A.P.


pa t p 1 d a

qa t q 1 d b

ra t r 1 d c

Subtract qa from pa .

p 1 q 1 d a b

p q d a b

a bd

p q

Subtract ra from qa .

q 1 r 1 d b c

q r d b c

b cd

q r

Equating both the values of d obtained, we get



a b b c

p q q r

a b q r b c p q

aq bq ar br bp bq cp cq

bp cp cq aq ar br 0

By rearranging terms we get

aq ar bp br cp cq 0

a q r b r p c p q 0

a q r b r p c p q 0

Therefore, q r a r p b p q c 0 is proved.

16. If 1 1 1 1 1 1

a ,b ,cb c c a a b

are in A.P., prove that a,b,c are in A.P.

Ans:Given 1 1 1 1 1 1

a ,b ,cb c c a a b

are in A.P.

Therefore,

1 1 1 1 1 1 1 1b a c b

c a b c a b c a

b a c a b c c a b b a c

ac bc ab ac

2 2 2 2 2 2 2 2b a b c a b a c c a c b b a b c

abc abc

2 2 2 2 2 2 2 2b a b c a b a c c a c b b a b c

2 2 2 2ab b a c b a a c b bc c b

ab b a c b a b a a c b c b bc c b

b a ab cb ca c b ac ab bc

b a c b

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

17. If a,b,c,d are in G.P., prove that n n n n n na b , b c , c d are in G.P

Ans: Given:

a,b,c and d are in G.P.



Therefore, 2b ac 2c bd

ad bc

To prove:

n n n n n na b , b c , c d are in G.P.

That is, 2

n n n n n nb c a b , c d

Then,

L.H.S 2

n nb c

2n n n 2nb 2b c c

n n

2 n n 2b 2b c c

n nn nac 2b c bd

n n n n n n n na c b c b c b d n n n n n n n na c b c a d b d

n n n n n nc a b d a b

n n n na b a d

R.H.S

Therefore,

2

n n n n n nb c a b c d

Therefore, n n n nb c , b c and n nc d are in G.P.

18. If a and b are the roots of 2

x 3x p 0 and c,d are roots of 2

x 12x q 0

, where a,b,c,d form a G.P. Prove that q p : q p 17 :15 .

Ans:Given: a and b are the roots of 2x 3x p 0 .

Therefore,

a b 3 and ab p …(1)

We also know that c and d are the roots of 2x 12x q 0 .

Therefore,

c d 12 and cd q …(2)

Also, a,b,c,d are in G.P.



Let us take 2a x,b xr,c xr and 3d xr .

We get from (1) and (2) that,

x xr 3

x 1 r 3

Also, 2 3xr xr 12

2xr 1 r 12

Divide both the equations obtained.

2xr 1 r 12

x 1 r 3

2r 4

r 2

3 3x 1

1 2 3

, when r 2 and

3 3x 3

1 2 1

, when r 2 .

Case I: 2ab x r 2 , 2 5cd x r 32 when r 2 and x 1 .

Therefore,

q p 32 2 34 17

q p 32 2 30 15

q p : q p 17 :15

Case II: 2ab x r 18 , 2 5cd x r 288 when r 2 and x 3 .

Therefore,

q p 288 18 306 17

q p 288 18 270 15

q p : q p 17 :15

Therefore, it is proved that q p : q p 17 :15 as we obtain the same for both the

cases.



19. The ratio of the A.M and G.M. of two positive numbers a and b is m :n .

Show that 2 2 2 2a :b m m n : m m n .

Ans:Let a and b be the two numbers.

The arithmetic mean, A.M a b

2

and the geometric mean, G.M ab


a b m

n2 ab

2 2

2

a b m

4 ab n

2

2

4abma b

n

2 abm

a bn

…(1)

Using the above equation in the identity 2 2

a b a b 4ab , we obtain

2 22

2

2 2

4ab m n4abma b 4ab

n n

2 22 ab m n

a bn

…(2)

Add equation (1) and (2)

2 22 ab2a m m n

n

2 2aba m m n

n

Substitute in (1) the value of a .

2 22 ab abb m m m n

n n

2 2ab abm m n

n n

2 2abm m n

n

Therefore,



2 2 2 2

2 22 2

abm m n m m na na : b

b ab m m nm m nn

Therefore, it is proved that 2 2 2 2a : b m m n : m m n .

20. If a,b,c are in A.P; b,c,d are in G.P. and 1 1 1

, ,c d e

are in A.P. Prove that a,c,e

are in G.P.

Ans:Given a,b,c are in A.P.

Therefore, b a c b

2b a c

a cb

2

…(1)

It is given b,c,d are in G.P.

Therefore, 2c bd 2c

db

…(2)

Also, 1 1 1

, ,c d e

are in A.P.

1 1 1 1

d c e d

2 1 1

d c e …(3)

We have to prove that a,c,e are in G.P.

That is, 2c ae .

Substitute (1) and (2) in (3).

2

2b 1 1

c c e

2

2 a c 1 1

c c e

2

a c e c

c ce



a c e c

c e

a c e e c c 2ae ce ec c

2c ae

Therefore, it is proved that a,c,e are in G.P.

21. Find the sum of the following series up to n terms:

i. 5 55 555 ...

Ans:Let nS 5 55 555... to n terms.

5

9 [9 99 999 ... to n terms]

5

9 [ 2 310 1 10 1 10 1 ... to n terms]

5

9 [( 2 310 10 10 ... to n terms) (1 1 to n terms)]

n10 10 15n

9 10 1

n10 10 15n

9 9

n50 5n10 1

81 9

Therefore, the sum of n terms of the given series is n50 5n10 1

81 9 .

ii. .6 .66 .666. ...

Ans:Let nS 0.6 0.66 0.666 to n terms.

6 [0.1 0.11 0.111 ... to n terms]

6

9 [0.9 0.99 0.999 ... to n terms]

6

9 [

2 3

1 1 11 1 1 ...

10 10 10

to n terms]



2

3 [(1 1 ... to n terms)

1

10 (

2

1 11

10 10 to n terms)]

n1

12 1 10

n13 10

110

n2 2 10n 1 10

3 30 9

n2 2n 1 10

3 27

Therefore, the sum of n terms of the given series is n2 2n 1 10

3 27

.

22. Find the th20 term of the series 2 4 4 6 6 8 ... n terms.

Ans:2 4 4 6 6 8 ... n is the given series,

Therefore the thn term 2

na 2n 2n 2 4n 4n

Then,

2

20a 4 20 4 20

4 400 80

1600 80

1680

Therefore, 1680 is the th20 term of the series.

23. Find the sum of the first n terms of the series: 3 7 13 21 31 ...

Ans:3 7 13 21 31 ... is the given series.

n 1 nS 3 7 13 21 31 ... a a

n 2 n 1 nS 3 7 13 21 ... a a a

Subtract both the equations.

n 1 n n 1 nS S 3 7 13 21 31 ... a a 3 7 13 21 31 ... a a

n n 1 nS S 3 7 3 13 7 21 13 ... a a a

n0 3 4 6 8 ... n 1 terms a

na 3 4 6 8 ... n 1 terms



n

n 1a 3 2 4 n 1 1 2

2

n 1

3 8 n 2 22

n 1

3 2n 42

3 n 1 n 2

23 n n 2

2n n 1 Therefore,

n n n n2

k

k 1 k 1 k 1 k 1

a k k 1

n n 1 2n 1 n n 1n

6 2

n 1 2n 1 3 n 1 6n

6

22n 3n 1 3n 3 36n

6

22n 6n 10n

6

2nn 3n 5

3

Therefore, the sum of n terms of the given series is 2nn 3n 5

3 .

24. If 1 2 3

S ,S ,S are the sum of first n natural numbers, their squares and their

cubes, respectively, show that 2

2 3 19S S 1 8S .

Ans:According to the conditions given in the question,

1

n n 1S

2



22

3

n n 1S

4

Therefore,

22

3 1

n n 1 8n n 1S 1 8S 1

4 2

22

2n n 11 4n 4n

4

222n n 1

2n 14

2

n n 1 2n 1

4

…(1)

And.

2

2

2 2

n n 1 2n 19S 9

6

29

n n 1 2n 136

2

n n 1 2n 1

4

…(2)

Therefore, we get 2

2 3 19S S 1 8S from (1) and (2).

25. Find the sum of the following series up to n terms: 3 3 3 3 3 3

1 1 2 1 2 3...

1 1 3 1 3 5

Ans:

2

3 3 3 3

n n 1

21 2 3 ... n

1 3 5 ... 2n 1 1 3 5 ... 2n 1

is the thn term of the given

series.

With first term a , last term 2n 1 and number of terms as n then 1,3,5... 2n 1

is an A.P.

Therefore,

2n1 3 5 ... 2n 1 2 1 n 1 2 n

2



Also,

2 22

2

n 2

n n 1 n 1 1 1 1a n n

4n 4 4 2 4

And, n n

2

n k

k 1 k 1

1 1 1S a K K

4 2 4

n n 1 2n 1 n n 11 1 1n

4 6 2 2 4

n n 1 2n 1 6 n 1 6

24

2n 2n 3n 1 6n 6 6

24

2n 2n 9n 13

24

Therefore, the sum of n terms of the given series

2n 2n 9n 13

24

.

26. Show that

22 2

2 2 2

1 2 2 3 ... n n 1 3n 5

1 2 2 3 ... n n 1 3n 1

.

Ans: 2 3 2n n 1 n 2n n is the thn term of the numerator.

2 3 2n n 1 n n is the thn term of the denominator.

n n3 2

22 2 k

k 1 k 1n n2 2 2

3 2

k

k 1 k 1

a K 2K K1 2 2 3 ... n n 1

1 2 2 3 ... n n 1a K K

…(1)

n

3 2

k 1

K 2K K

22n n 1 2n n 1 2n 1 n n 1

4 6 2

n n 1 n n 1 22n 1 1

2 2 3



2n n 1 3n 3n 8n 4 6

2 6

2n n 13n 11n 10

12

2n n 13n 6n 5n 10

12

n n 13n n 2 5 n 2

12

n n 1 n 2 3n 5

12

…(2)

And,

n

3 2

k 1

K K

22n n 1 n n 1 2n 1

4 6

n n 1 n n 1 2n 1

2 2 3

2n n 1 3n 3n 4n 2

2 6

2n n 13n 7n 2

12

2n n 13n 6n n 2

12

n n 13n n 2 1 n 2

12

n n 1 n 2 3n 1

12

…(3)

Substituting (2) and (3) in (1), we get

22 2

2 2 2

n n 1 n 2 3n 51 2 2 3 ... n n 1 12

n n 1 n 2 3n 11 2 2 3 ... n n 1

12



n n 1 n 2 3n 5

n n 1 n 2 3n 1

3n 5

3n 1

Therefore,

22 2

2 2 2

1 2 2 3 ... n n 1 3n 5

1 2 2 3 ... n n 1 3n 1

is proved.

27. A farmer buys a used tractor for Rs.12000 . He pays Rs.6000 cash and agrees

to pay the balance in annual installments of Rs.500 plus 12% interest on the

unpaid amount. How much will be the tractor cost him?

Ans:It is given that Rs.6000 is paid in cash by the farmer.

Therefore, the unpaid amount is given by

Rs.12000 Rs.6000Rs.6000

According to the conditions given in the question, the interest to be paid annually by

the farmer is

12% of 6000 , 12% of 5500 , 12% of 5000...12% of 500

Therefore, the total interest to be paid by the farmer

12% of 6000 12% of 5500 12% of 5000 ... 12% of 500

12% of 6000 5500 5000 ... 500

12% of 500 1000 1500 ... 6000

With both the first term and common difference equal to 500 , the series

500,1000,1500...6000 is an A.P.

Let n be the number of terms of the A.P.

Therefore,

6000 500 n 1 500

1 n 1 12

n 12 Therefore, the sum of the given A.P.

12

2 500 12 1 5002

6 1000 5500

6 6500

39000 Therefore, the total interest to be paid by the farmer



12% of 500 1000 1500 ... 6000

12% of Rs.39000

Rs. 4680

Therefore, the total cost of tractor

(Rs.12000Rs. 4680 )

Rs.16680

Therefore, the total cost of the tractor is Rs.16680 .

28. Shamshad Ali buys a scooter for Rs. 22000 . He pays Rs.4000 cash and agrees

to pay the balance in annual installment of Rs.1000 plus 10% interest on the

unpaid amount. How much will the scooter cost him?

Ans:It is given that for Rs.22000 Shamshad Ali buys a scooter and Rs. 4000 is paid

in cash.

Therefore, the unpaid amount is given by

Rs.22000 Rs.4000Rs.18000

According to the conditions given in the question, the interest to be paid annually

is

10% of 18000 , 10% of 17000 , 10% of 16000...10% of 1000

Therefore, the total interest to be paid by the farmer

10% of 18000 10% of 17000 10% of 16000 ... 10% of 1000

10% of 18000 17000 16000 ... 1000

10% of 1000 2000 3000 ... 18000

With both the first term and common difference equal to 1000 , the series

1000,2000,3000...18000 is an A.P.

Let n be the number of terms of the A.P.

Therefore,

18000 1000 n 1 1000

1 n 1 18

n 18 Therefore, the sum of the given A.P.

18

2 1000 18 1 10002

9 2000 17000

9 19000

171000 Therefore, the total interest to be paid



10% of 18000 17000 16000 ... 1000

10% of Rs.171000

Rs.17100

Therefore, the total cost of scooter

(Rs.22000Rs.17100)

Rs.39100

Therefore, the total cost of the scooter is Rs.39100 .

29. A person writes a letter to four of his friends. He asks each one of them to

copy the letter and mail to four different persons with instruction that they move

the chain similarly. Assuming that the chain is not broken and that it costs 50

paise to mail one letter. Find the amount spent on the postage when th8 set of

letter is mailed.

Ans:2 84,4 ,...4 is the number of letters mailed and it forms a G.P.

The first term a 4 , the common ratio r 4 and the number of terms n 8 of the

G.P.

We know that the sum of n terms of a G.P. is

n

n

a r 1S

r 1

Therefore,

8

8

4 4 1S

4 1

4 65536 1

3

4 65535

3

4 21845

87380 50 paisa is the cost to mail one letter.

Therefore,

Cost of mailing 87380 letters Rs.50

87380100

Rs.43690

Therefore, Rs. 43690 is the amount spent when th8 set of letter is mailed.



30. A man deposited Rs.10000 in a bank at the rate of 5% simple interest

annually. Find the amount in th15 year since he deposited the amount and also

calculate the total amount after 20 years.

Ans:Rs.10000 is deposited by the man in a bank at the rate of 5% simple interest

annually

5

100 Rs.10000 Rs.500

Therefore,

10000 500 500 ... 500 is the interest in th15 year. (500 is 14 added times)

Therefore, the amount in th15 year

Rs.10000 14 Rs.500

Rs.10000Rs.7000

Rs.17000

Rs.10000 500 500 ... 500 is the amount after 20 years. (500 is 20 added

times)

Therefore, the amount after 20 years

Rs.10000 20 Rs.500

Rs.10000Rs.10000

Rs.20000

The total amount after 20 years is Rs.20000 .

31. A manufacturer reckons that the value of a machine, which costs him Rs.

15625 , will depreciate each year by 20%. Find the estimated value at the end of

5 years.

Ans:The cost of the machine is Rs.15625 .

Every year machine depreciates by 20%.

Therefore, 80% of the original cost ,i.e., 4

5 of the original cost is its value after every

year.

Therefore, the value at the end of 5 years

4 4 415626 ...

5 5 5

5 1024

5120 Therefore, Rs.5120 is the value of the machine at the end of 5 years.



32. 150 workers were engaged to finish a job in a certain number of days. 4

workers dropped out on second day, 4 more workers dropped out on third day

and so on. It took 8 more days to finish the work. Find the number of days in

which the work was completed.

Ans:Let the number of days in which 150 workers finish the work be x .


150x 150 146 142 ... x 8 terms

With first term a 146 , common difference d 4 and number of turns as x 8 ,

the series 150 146 142 ... x 8 terms is an A.P.

x 8150x 2 150 x 8 1 4

2

150x x 8 150 x 7 2

150x x 8 150 2x 14

150x x 8 136 2x

75x x 8 68 x 275x 68x x 544 8x

2x 75x 60x 544 0 2x 15x 544 0 2x 32x 17x 544 0

x x 32 17 x 32 0

x 17 x 32 0

x 17 or x 32

We know that x cannot be negative.

So, x 17 .

Therefore, 17 is the number of days in which the work was completed. Then the

required number of days 17 8 25 .


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NCERT Solutions for Class 11 Maths Chapter 9 – Sequences ...

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