8 . SEQUENCES & SERIES

CBSE Class 11 Maths Chapter 8 . SEQUENCES & SERIES

Chapter 8. SEQUENCES & SERIES

Class 11 Maths Chapter 8. Sequences and Series Exercise 8.1 Solutions :

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

1. 2. 3. 4. 5. 6.

Solution :

1. We have,

2. We have,

3. We have,

4. We have,

5. We have,

6. We have,

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

7. 8. 9. 10.

Solution: 7. We have,

8. We have,

9. We have,

10. We have,

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

11. for all

12.

13. .

Solution :

11. We have, for all

The series is :

12. We have,

The series is :

13. We have, .

The series is :

14. The Fibonacci sequence is defined by and . Find , for

Solution : Here ,

Again, , For

and

Class 11 Maths Chapter 9. Sequences and Series Exercise 9.2 Solutions :

1. Find the 20th and nth terms of the G.P.

Solution : Here , and

We have,

Again,

2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Solution : let , be the first term of a G.P.

Here, and

We know that ,

Now,

3. The 5th, 8th and 11th terms of a G.P. are and s , respectively. Show that .

Solution : Here, , and

We know that ,

Again,

And

LHS :

RHS

4. The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.

Solution : let be the ratio of a G.P. and .

A/Q,

Again ,

5. Which term of the following sequences:
(a) is 128 ?

(b) is 729 ?

Solution : (a) is 128

Here, , and

We know that ,

Therefore, the number is 12 .

(b) is 729 .

Here, , and

We know that ,

(c)

Here, , and

We know that ,

Therefore, the number of the term is 8 .

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

Solution : We have,

Therefore, the value of is .
Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:
7. 0.15, 0.015, 0.0015, ... 20 terms.

Solution : Here, and

We know that ,

8. terms.

Solution : Here, and

We know that ,

9. terms (if ) .

Solution : Here, first term and

10. terms (if ).

Solution : Here , and

We know that

11. Evaluate :

Solution : We have,

12. The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.

Solution: let , the three terms are : and .

A/Q,

and

Putting in eq. (i) , we get

For :

, and

Therefore, the G.P. is : , 1 and .

For :

, ,

Therefore, the G.P. is : , 1 and .

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

Solution : Here, , and

We know that ,

Therefore, the number of terms is 4 .

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.

Solution : let, the G.P. is : .

A/Q,

Putting in eq. (i) , we get

We have,

15. Given a G.P. with and 7th term 64, determine .

Solution : let , be the common ratio of a G.P .

Here, and

A/Q,

We have,

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.

Solution : let the G.P. is : .

A/Q,

Again,

(impossible) or

If , then

[From (i)]

Therefore, the G.P. is :

I.e., 4 , – 8 , 16 , ……………. .

If , then [From (i)]

Therefore, the G.P. is :

i.e.,

17. If the 4th, 10th and 16th terms of a G.P. are and respectively. Prove that are in G.P.

Solution : let and are first term and common ratio of the G.P. respectively .

A/Q,

Therefore, are in G.P.

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

Solution : Let to terms .

19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,16, 32 and 128, 32, 8, 2, .

Solution : We have,

i.e. , 256 , 128 , 64 , 32 , 16 .

let

Here , , and

We have,

20. Show that the products of the corresponding terms of the sequences and form a G.P, and find the common ratio.

Solution : The first sequences are :

The second sequences are :

A/Q,

The common ratio

21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Solution : let ,the G.P. are

A/Q,

And

Putting in eq. (i) , we get

Therefore, the G.P. is :

i.e., 3 , – 6 , 12 , – 24 , ………….

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

Solution : Let and R be the first term and common ratio of the G.P. respectively .

A/Q,

Again ,

And

Proved .
23. If the first and the nth term of a G.P. are and , respectively, and if is the product of terms, prove that .

Solution : let , The G.P is :

Here, first term

and

A/Q, terms

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 .

Solution : let the GP is : .

For the sum of first term :

For the sum of to term :

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

Solution : Since, and are in G.P .

Let,

, and

R.H.S :

LHS Proved .

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

Solution : Let , the two number are and , then 3 , , , 81 are in GP .

We have,

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

Solution : We have,

[ ]

Therefore, the value of n is .

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

Solution : let and be the two numbers .

Geometric mean

A/Q,

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

Solution : Let ,the numbers are and .

A/Q,

And

The quadratic equation of the roots are and ,then

Apply quadratic formula , We have

Proved.

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 hour, hour and hour ?

Solution : Here, and common ratio

We have ,

Again ,

And

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?

Solution : Here, , and

We know that,

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

Solution : let a and b are the roots of the quadratic equation .

A/Q,

And

The quadric equation is

Class 11 Maths Chapter 9. Sequences and Series Miscellaneous Exercise Solutions :

1. If is a function satisfying for all such that and , find the value of .

Solution : We have,

Given,

If , then

If and , then

Now ,

Here, , and

We know that ,

2. 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.

Solution : let , the G.P. is : .

Here, , and

We know that ,

We have,

3. 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.

Solution: let be the common ratio . First term .

A/Q,

(Impossible ) or

Therefore, the common ratio is ± 3 .

4. The sum of 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.

Solution : Let the three numbers in the geometric progression be .

and the common ratio be .

A/Q,

Subtracting 1, 7, and 21 in order gives an A.P : and .

If , then

For : and , then . i.e., 8 , 16 , 32

For : and , then

i.e., 32 , 16 , 8

So, the three numbers in the geometric progression are 8, 16, and 32 or 32 , 16 , 8 .

5. 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.

Solution : Let and be the first term and common ratio of the G.P. respectively .

So, the GP are : .

A/Q,

Therefore, the common ratio is 4 .

6. If , then show that and are in G.P.

Solution : We have,

Therefore, and are in G.P .

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

Solution : let the G.P is : .

A/Q

Again ,

And

LHS :

RHS

Proved .

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

Solution : Since, are in G.P .

Let ,

We have,

LHS :

RHS .

Therefore, , , are in G.P.

Proved .

9. If and are the roots of and are roots of where form a G.P. Prove that .

Solution : Given, are in G.P.

Let,

, and

Since, and are the roots of , then

And

Again, and are roots of , then

And

Putting in eq. (i) , we get

, and

,and

Proved .

Putting in eq. (i) , we get

, and

,and

From (ii) , we get

From (iv) , we get

Proved .

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

Solution : Given and be the two numbers .

Arithmetic mean

Geometric mean

A/Q,

11. Find the sum of the following series up to terms:
(i) (ii)

Solution : (i) Let , to terms .

(ii)

Let , to terms .

12. Find the term of the series terms.

Solution : Given, the series terms.

Let , terms

13. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

Solution : The farmer buys a used tractor for Rs 12000.

He pays Rs 6000 in cash.

So, the remaining amount to be paid is Rs 12000 – Rs 6000 = Rs 6000.

Now, the farmer agrees to pay the balance in annual installments of Rs 500 plus 12% interest on the unpaid amount.

Here, P = 6000 , R = 12 % and T = 1

We know that ,

Interest for first installments

Unpaid amount = 6000 – 500 = Rs 5500

Interest for second installments

Unpaid amount = 5500 – 500 = Rs 5000

Interest for third installments

Total interest = 720 + 660 + 600 + ………….+ 12 terms

Here, , ,and

Therefore, the total amount .
14. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Solution: Shamshad Ali buys a scooter for Rs 22000.

He pays Rs 4000 in cash.

So, the remaining amount to be paid is Rs 22000 – Rs 4000 = Rs 18000.

Now, he agrees to pay the balance in annual installments of Rs 1000 plus 10% interest on the unpaid amount.

Here, P = 18000 , R = 12 % and T = 1

We know that ,

Interest for first installments

Unpaid amount = 18000 – 1000 = Rs 17000

Interest for second installments

Unpaid amount = 17000 – 1000 = Rs 16000

Interest for third installments

Total interest = 1800 + 1700 + 1600 + ………….+ 12 terms

Here, , , and

Therefore, the total amount .

15. 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 8th set of letter is mailed.

Solution : The person writes a letter and sends it to four friends. Each friend is instructed to copy the letter and mail it to four different persons

In the first set, there are 4 × 1 = 4 letters.

In the secondset, there are 4 × 4 = 16 letters.

In the third set, there are 4 × 16 = 64 letters.

The G.P. is : 4 , 16 , 64 , ……. .

Here, , , and

We know that ,

Therefore, the total postage cost .

16. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in year since he deposited the amount and also calculate the total amount after 20 years.

Solution : Here , P = 10000 , R = 5 % , T = 1

We know that ,

Interest for 1 years

Total amount for 1 year = 10000 + 500 = Rs 10500

Interest for 2 years

Total amount for 2 year = 10000 + 1000 = Rs 11000

Interest for 3 years

Total amount for 3 year = 10000 + 1500 = Rs 11500

The amount (in Rs) are : 10000 , 10500 , 11000 , 11500 , …..

Here, and

Therefore, the amount after 15 years is Rs 17000 .

Here, and

Therefore, the amount after 20 years is Rs 19500 .

17. 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.

Solution : We know that ,

[ Depreciation is the decrease in the value of an asset.Top of Form ]

The initial value (cost of the machine) is Rs. 15625, the rate of depreciation is 20%, and the number of years is 5.

Here, P = 15625 , R = 20% and n = 5 years

The estimated value at the end of 5 years

18. 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.

Solution : Let be the number of days in which 150 workers to finish the work .

The number of workers engaged on the first day is 150 and then the number decreases by 4 each day.

Total work

The number of workers on the second day is 150 – 4 = 146 .

The number of workers on the third day is 146 – 4 = 142, and so on.

The AP’s is : 150 , 146 , 142 , …………. .

Here,

It is given that it took 8 more days to finish the work, so the total number of days becomes .

A/Q,

(impossible)

Therefore, the total number of days is 25 ( = 17 + 8) .