cbse ncert solutions for class 11 mathematics chapter 09 · 2019. 11. 27. · back of chapter...

Class–XI–CBSE-Mathematics Sequences and Series

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Exercise 9.1

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

Hint: Put .

Solution:

Solution step 1:

Putting , and we get

So, the required first five terms are and

2. Write the first five terms of the sequences whose term is

Hint: Put .

Solution:

Solution step 1:

Putting , we get

Back of Chapter Questions

CBSE NCERT Solutions for Class 11 Mathematics Chapter 09



So, the required first five terms are


Hint: Put .

Solution:

Solution step 1:

Putting , we get

So, the required first five terms are and


Hint: Put .

Solution:

Solution step 1: Putting , we get



So, the required first five terms are. .


Hint: Put .

Solution:


So, the required first five terms are , and


Hint: Put .

Solution:




So, the required first five terms are .

7. Find the and term in the following sequence whose term is

Hint: Put .

Solution:


Substituting , we get

8. Find the term in the following sequence whose term is

Hint: Put .

Solution:



Hint: Put .

Solution:



Hint: Put .

Solution:

Solution step 1: Substituting , we obtain



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

for all

Hint: Put and use the previous term wisely.

Solution:

Solution step 1: for all

Hence, the first five terms of the sequence are and .

So the series is



Solution:

Solution step 1:



Hence, the first five terms of the sequence are

So the series is



Solution:

Solution step 1:

Hence, the first five terms of the sequence are and

So the series is

14. The Fibonacci sequence is defined by

Find , for


Solution:

Solution step 1:

For



For

For

For

For

Exercise 9.2

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

Hint: and

Solution:

Solution step 1: The odd integers from to are It is in A.P.

So, first term,

Difference between 2 numbers,

Here,

So, the sum of odd numbers from to is



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

Hint: and

Solution:

Solution step 1: So the series is .

So

So, the sum of all natural numbers lying between and , which are multiples of , is .

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

Show that term is

Hint:

Solution:

Solution step 1:First term

Let be the difference betweentwo numbers of the A.P.

So, the A. P. is .

Sum of first five terms

Sum of next five terms



[by question]

So, the of the A.P. is

4. How many terms of the A.P. are needed to give the sum ?

Hint:

Solution:

Solution step 1: Assume number of terms first term, and common difference

Here, [By question]

Sum of all term in A.P.=

[ ]

or



5. In an A.P., if term is and term is , prove that the sum of first terms is ,

where

Solution:

Solution step 1:

Using the above formula term will be

term

term

By subtracting from , we get

Putting the value of in , we get

So, the sum of first terms of the A.P. is. .



6. If the sum of a certain number of terms of the A.P. is Find the last term

Hint: and

Solution:

Solution step 1: Sum of all term of A.P. is 116.

And, and

Putting the values in above formula we get

or

n should be a integer, so

Last term

So, the last term of the A.P. is .

7. Find the sum to terms of the A.P., whose term is

Hint: and

Solution:

Solution step 1: term a



So

Comparing the coefficient of , we get

8. If the sum of terms of an A.P. is , where and are constants find the common

difference.

Hint: Sum of n number in A.P.

Solution:

Solution step 1: Sum of all number in A.P.:

So [By question]

Comparing the coefficients of on both sides, we get

So, the common difference of the A.P. is

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

their terms.



Hint: The general term in A.P.

Solution:

Solution step 1: Assume first terms and be as common difference of A.P.

Putting in , we get

By equation and , we get

So, the ratio of term of both the A.P.S is

10. If the sum of first terms of an A.P. is equal to the sum of the first terms, then find the sum of the

first terms.

Hint: The general term in A.P. , Sum of n number in

A.P.

Solution:

Solution step 1: Assume a be as first term and d be as common difference of A.P.

The general term in A.P.



Using the above formula and are

Sum of n number in A.P.

Using the above formula

[ By question]

So, the sum of the first terms of the A.P. is

11. Sum of the first and terms of an A.P. are and , respectively.

Prove that



Hint: The general term in A.P. , Sum of n number in

A.P.

Solution:



According to the above formula and are:

[ by question]

[ by question]

[ by question]

Subtracting from , we get

Subtracting from , we get



By equation (iv) and (v) we get

[Dividing both sides by ]

Hence proved.

12. The ratio of the sums of and terms of an A.P. is . Show that the ratio of and term

is .

Hint: Sum of n number in A.P. ,The general term in A.P.

Solution:



Using the above formula we get:

Putting and in , we get





From and , we get

Hence proved.

13. If the sum of terms of an A.P. is and its term is find the value of


Solution:






Comparing the coefficient of on both sides, we get



Comparing the coefficient of on both sides, we obtain

So, from equation , we get

So, the value of is .

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


Solution:

Solution step 1: Assume and be five numbers between and .

So, a



So, the five numbers are and



15. If is the A.M. between and , then find the value of

Hint: A. M. of and

Solution:

Solution step 1: A. M. of and

[ ]

So, the required value of n is 1.

16. Between and , numbers have been inserted in such a way that the resulting sequence is an

A.P. and the ratio of and numbers is Find the value of


Solution:

Solution step 1: Assume , be numbers such that is an A.P.

So,





According to the given condition,

So, the required value of m is 14.

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

every month, what amount he will pay in the installment?


Solution:

Solution step 1: The first installment of the loan is₹100 and the second installment of the loan is

₹ and so on.



So the A. P. is

So,



So, the installment is Rs .

18. The difference between any two consecutive interior angles of a polygon is . If the smallest angle

is , find the number of the sides of the polygon.

Hint: Sum of all angles of a polygon with sides is ,Sum of n number in

A.P.

Solution:

Solution step 1: The angles of the polygon will form an A.P. So the A.P. will be

Sum of all angles of a polygon with sides is


By using the equation (i) and (ii) we get:



or

Exercise 9.3

1. Find the and terms of the G.P.

Hint:

Solution:

Solution step 1: G.P. is

So, first term , common ratio

Using the above formula and will be:

2. Find the term of a G.P. whose term is 192 and the common ratio is

Hint:

Solution:

Solution step 1: Assume a be the first term and r be the common ratio of the G.P.

So common ratio,



By using the above formula we get:

So the required term is .

3. The and terms of a G.P. are and , respectively. Show that

Hint:

Solution:


By using the above equation we get:

Dividing equation by , we get


From equation and , we get



Hence proved.

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

Hint:

Solution:


[ By question]


[ By question]

So, the term of the G.P. is

5. Which term of the following sequences:

, is ?

Hint:

Solution:

(a) The given sequence is , is 128?


So, and



term= . [ By question]


So, the term is

, is 729?

(b) The given sequence is

Hint:

Solution:


So, and

term= . [ By question]




So, the term is

, is ?

(c) The given sequence is

Hint:

Solution:


So,

term = . [ By question]


So, the term is. .

6. For what values of , the numbers are in G.P?

Hint:



Solution:

Solution step 1: Numbers are [ By question]

Common ratio

Also, common ratio [ common ratio will be same for each term in G.P.]

So, for , the given numbers will be in G.P.

7. Find the sum to terms the geometric progression

Hint:

Solution:

Solution step 1: The given G.P. is

So, and

By the above formula we get:



8. Find the sum to terms in the geometric progression

Hint:

Solution:


So, and



Hint:

Solution:


So, first term and Common ratio





Hint:

Solution:


So, and


11. Evaluate

Hint:

Solution:

Solution step 1:

3, forms a G.P.

So,




Putting the value in equation , we get:

12. The sum of first three terms of a G.P. is and their product is . Find the common ratio and the

terms.

Hint: Assume be the first three terms of the G.P.

Solution:

Solution step 1: Take the be the terms of the G.P.

[ By question]

[ By question]

From we get

(Taking only real root)

Putting in equation we get

or

So the 3 term of G.P. are .



13. How many terms of G.P. are needed to give the sum 120?

Hint:

Solution:


So, and


[ By question]

So the four terms are required to get the sum as

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

first term, the common ratio and the sum to terms of the G.P.

Hint: Assume the G.P, is a

Solution:

Solution step 1: Assume the G.P. be

and [ By question]

Dividing equation by we get



Putting in ,we get:

By using the above formula we get;

15. Given a G.P. with and term , determine

Hint: and

Solution:

Solution step 1: Assume be the common ratio of the G.P.

and [ By question]

be the common ratio of the G.P.





So the required is .

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

Hint: and

Solution:

Solution step 1: Assume a be the first term and r be the common ration of A.P.

By using the above equation we get;



From equation we get:

for

Also, for

So, the required G.P. is or

17. If the and terms of a G.P. are and , respectively. Prove that are in G.P.

Hint: and

Solution:







So, are in G. P.

18. Find the sum to terms of the sequence,

Hint:

Solution:

Solution step 1: The given sequence is

This sequence is not a G.P. But, we can change itinto G.P. as to

terms

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

Hint:



Solution:

Solution step 1: Required sum

So the series is in G.P.

So term, and Common ratio,


Required sum

20. Show that the products of the corresponding terms of the sequences form

a G.P, and find the common ratio.

Hint: Common ratio

Solution:

Solution step 1: By question we get .

So the common ratio is same for all term, the series is in G.P.

21. Find four numbers forming a geometric progression which third term is greater than the first term

by , and the second term is greater than the by .

Hint:

Solution:





,

[ By question]

[ By question]

From equation and we get


Putting the value of in we get

So, the first four numbers of the G.P. are and i.e., and

22. If and terms of a G.P. are and , respectively. Prove that

Hint:

Solution:

Solution step 1: Assume be the first term and be the common ratio of the G.P.

By using the above equation we get:

[ By question]

[ By question]



[ By question]

Hence proved.

23. If the first and the term of a G.P. are ad , respectively, and if is the product of terms,

prove that

Hint:

Solution:

Solution step 1: Assume the first term of the G.P is a and the last term is

Take the G.P. is , where is the common ratio.

[ By question]

Product of terms

So, is an A. P.



Hence proved.

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

term is

Hint: and Sum of first terms

Solution:

Solution step 1: Assume a be the first term and be the common ratio of the G.P.

Sum of first terms


Since there are terms from to term, Sum of terms from to term

So, required ratio

Hence proved.

25. If and are in G.P. show that:

Hint:

Solution:

Solution step 1: are in G.P.[by the question]

So

[by the question]



R.H.S.

Hence proved.

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

Hint:

Solution:

Solution step 1: Assume and be two numbers between and so that the series, ,

forms a G.P.

Assume a be the first term and be the common ratio of the G.P.

(consider only real roots)

For



So, the required two numbers are and .

27. Find the value of so that may be the geometric mean between and

Hint: Middle term of and is

Solution:

Solution step 1: G.M.


[ By question]

Squaring both sides, we get

So the required number n is

28. The sum of two numbers is times their geometric mean, show that number are in the ratio

Hint: Middle term of and is



Solution:

Solution step 1: Assume the two numbers be and

G.M.


[ By question]

Also,

Adding and , we get:

Putting the value of a in we get:

So, the required ratio is

29. If A and be A.M. and G.M., respectively between two positive numbers, prove that the numbers

are

Hint: and

Solution:

Solution step 1: Assume two numbers be and



From equation and , we get:

a

Substituting the value of a and from (iii) and (iv) in the (v) we get

From equation and , we get

Putting the value of a in , we get:

So the two numbers are .

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

culture originally, how many bacteria will be present at the end of hour, hour and hour?

Hint:

Solution:

Solution step 1: Bacteria doubles every hour so the number of bacteria after every hour will form a

G.P.

So, and


So, the number of bacteria at the end of hour will be 120.

The number of bacteria at the end of hour will be



So, number of bacteria at the end of hour will be

31. What will ₹ amounts to in years after its deposit in a bank which pays annual interest rate of

compounded annually?

Hint:

Solution:

Solution step 1: At the end of first year, amount ₹.

At the end of year, amount

At the end of year, amount and so on

Amount at the end of years

32. If A.M. and G.M. of roots of a quadratic equation are and respectively, then obtain the quadratic

equation.

Hint: and

Solution:

Solution step 1: Assume the root of the quadratic equation be and

Formula of quadric equation is given by:

(Sum of roots) (Product of roots)

By equation (ii) and (iii) we get:

So the quadratic equation is



Exercise 9.4

1. Find the sum to terms of the series

Hint:

Solution:

Solution step 1: So the series is term,


Hint:

Solution:





Hint:

Solution:





Hint:

Solution:

Solution step 1: So the series is

term,

Adding the above terms we get:




Hint:

Solution:


term is


Hint:

Solution:





Hint:

Solution:




8. Find the sum to terms of the series whose term is given by .

Hint:

Solution:

Solution step 1:



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

Hint:

Solution:

Solution step 1:

Take

So the above series is in G.P. where and .

By using equation (i) and (ii) we get:

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

Hint:

Solution:

Solution step 1:



Miscellaneous Exercise

1. Show that the sum of and terms of an A.P. is equal to twice the term.

Hint:

Solution:

Solution step 1: Assume be the first term and the common difference of the A.P.


[ By question]

So, the sum of and terms of an A.P. is equal to twice the term.



2. If the sum of three numbers in , is 24 and their product is , find the numbers.

Hint: Assume the numbers are , and

Solution:

Solution step 1: Assume the three numbers in . be , and

[by question]

[by question]

So, when , the numbers are and and when the numbers are and

So, the three numbers are and

3. Let the sum of terms of an A.P. be and , respectively, show that

Hint:

Solution:

Solution step 1: Assume and be the first term and d is the common difference of the A.P.




From equation (i) and (ii) we get:

Hence proved.

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

Hint for Overall:

Solution:

Solution Step1: The numbers lying between and which are divisible by are

First term, ,Last term, ,Common difference,


[By question]



So, the required sum is 8729.

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

Hint:

Solution:

Solution step 1: The integers from to , which are divisible by are

So



The integers from to which are divisible by are

This forms an A.P. with both the first term and common difference equal to





The integers, which are divisible by both and are This also forms an A.P. with both

the first term and common difference equal to



Required sum

Thus, the sum of the integers from 1 to 100, which are divisible by or is

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

Hint:

Solution:

Solution step 1: The two‐digit numbers, which when divided by , yield as remainder, are

So and

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





So, the required sum is

7. If is a function satisfying for all , such that and

, find the value of

Hint:

Solution:

Solution step 1: By question:

for all

Taking in (1),

we take

Similarly,

that is forms a G.P.

So and .




[By question]

So, the value of is .

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

respectively. Find the last term and the number of terms.

Hint:

Solution:

Solution step 1: the sum of terms of the G.P. be

It is given that the first term a is 5 and common ratio is

Last term of the G. term

So the last term of the is

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

G.P.

Hint:



Solution:

Solution step 1: Assume abe the first term and be the common ratio of the G.P.

(consider only real roots)

So, the common ratio of the G.P. is

10. The sum of three numbers in . is . If we subtract from these numbers that order, we

obtain an arithmetic progression. Find the numbers.

Hint: Assume 3 terms of G.P. are , and

Solution:

Solution step 1: Assume the three numbers in G.P. be , and

[By question]

forms an A.P.

From equation (i) and (ii) we get



So

So, when , the three numbers in . are and and if, , the three numbers in

. are and

Both cases the required terms are .

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

occupying odd places, then find its common ratio.

Hint:

Solution:

Solution step 1: Let the G.P. be

Number of terms

[By question]

So the G.P. be

So, the common ratio of the G.P. is

12. The sum of the first four terms of an A.P. is . The sum of the last four terms is . If its first term

is , then find the number of terms.

Hint:

Solution:

Solution step 1: Assume the first four A.P. be



So sum of first four terms

So sum of last four terms

[by question]

So, the number of terms of the A.P. is

13. If then show that and are in G.P.

Hint: check the given in question carefully.

Solution:

Solution step 1: [by question]

[by question]



From equation and we get

So, and are in G.P.

14. Let be the sum, the product and the sum of reciprocals of terms in a G.P. Prove that

Hint: assume the n terms of G.P. are .

Solution:

Solution step 1: Assume the G.P. be

[By question]

[By question]

[By question]



15. The and terms of an A.P. are respectively. Show

that

Hint:

Solution:

Solution step 1: Let be the first term and d be the common difference of the A.P.

The term of an A.P.

By using the above formula we get and .

Subtracting equation from , we get

Subtracting equation from we get

By equation (iv) and (v) we get



Hence proved.

16. If a are in A.P., prove that are in A. P

Hint:

Solution:

Solution step 1: It is given the a are in A. P.

Thus, , and are in A.P.

17. If are in G.P, prove that are in G.P.

Hint: a, b, c will be in G.P. if .

Solution:

Solution step 1: , and are in G.P so:



So

Taking L. H.S.

R.H.S.

Hence proved.

18. If and are the roots of and , are roots of where ,

form a G.P.

Prove that

Hint: properties of quadric equation

Solution:

Solution step 1: a and are the roots of

and [By the properties of quadric equation]

Also, and are the roots of

and [By the properties of quadric equation]

are in G.P.[by question]

Assume From and

we get



On dividing, we get

When

When

Case I:When and

i.e.,

Case II:When

So,

So, in both the cases, we obtain 17.15

Hence proved.

19. The ratio of the A.M and G.M. of two positive numbers a and , is . Show

that

Hint: A.M and G.M.

Solution:

Solution step 1: Assume the two numbers be and

A.M and G.M.

[By question]



Using this the identity , we get

Adding and , we get

Putting the value of a in we get

So,

20. If are in A.P; are in G.P and are in A.P. prove that are in G.P.

Hint: Common ratio in G.P. and common difference in A.P. same for each term.



Solution:

Solution step 1: So are in A.P.

And , are in G.P.

Also, are in A.P.

We have to prove that

From equation , we get

From equation we get

Putting these values in , we get

So, , and are in G.P.

Hence proved.



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

(i) (ii)

Hint:

Solution:

Solution step 1: (i)

Take to terms

(ii)

Hint:

Solution:

Solution step 1: Take to terms



22. Find the term of the series terms.

Hint:

Solution:

Solution step 1: The given series is terms

term

So, the term of the series is 1680.

23. Find the sum of the first terms of the series:

Hint:

Solution:

Solution step 1: The given series is

On subtracting both the equations, we get



24. If are the sum of first natural numbers, their squares and their cubes, respectively, show

that

Hint:

Solution:

Solution step 1: By the question



Here,

Also,

Thus, from and , we obtain

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

Hint:

Solution:

Solution step 1: The term of the given series is

So, is an A.P. with first term a, last term and number of terms as



26. Show that

Hint:

Solution:

term of the numerator

term of the denominator

Here,



Also,

From and we get

So, the given result 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?

Hint: Total interest

Solution:

Solution step 1: The farmer pays Rs 6000 in cash [By question]

Then unpaid amount According to the given condition, the interest

paid annually is

So, total interest to be paid



Now, the series is an A.P. with both the firstterm and common difference

equal to 500.

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

Sum of the A.P


of of 4680

So, cost of tractor

28. Shams had 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?

Hint: Sum of terms of a G.P. is

Solution:

Solution step 1: Shamshad Ali buys a scooter for 22000 and pays 4000 in cash.[By question]

Unpaid amount

the interest paid annually is[By question]

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


of of of of 1000

of

of

Here, forms an A. P. with first term is 1000 and common difference is

1000.

So the number of terms be



Total interest paid of

of 17100

Cost of scooter

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 set of letter is mailed.

Hint: Sum of terms of a G.P. is

Solution:

Solution step 1: The numbers of letters mailed forms a

First term

Common ratio

Number of terms

Sum of terms of a G.P. is

By using the above formula

It is given that the cost to mail one letter is paisa.

Cost of mailing 87380 letters

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



30. A man deposited ₹ 10000 in a bank at the rate of 5% simple interest annually. Find the amount

in15th year since he deposited the amount and also calculate the total amount after 20 years.

Hint: Check the given in question wisely.

Solution:

Solution step 1: So the man deposited ₹ 10000 in a bank at the rate of 5%

simple interest annually.

Interest first year

Amount in15th year

Amount after 20 years

31. A manufacturer reckons that the value of a machine, which costs him , will depreciate each

year by 20%. Find the estimated value at the end of 5 years.

Hint: Use the given in question wisely.

Solution:

Solution step 1: Cost of machine

Machine depreciates by 20% every year.

Its value after every year is 80% of the original cost i.e., of the original cost.

Value at the end of 5 years

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

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.



Hint:

Solution:

Solution step 1: Assume the number of days in which 150 workers finish the work is x.

terms[By question]

The series terms is an A.P.

So the first term and common difference .

We consider only positive value of x so

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

So required number of days

cbse ncert solutions for class 11 mathematics chapter 09 · 2019. 11. 27. · back of chapter...

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