# NCERT Solutions class-11 Maths Miscellaneous Part-1

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Miscellaneous Exercise

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

Ans. Here,  …..(i)

and   …..(ii)

To prove:

Adding eq. (i) and (ii), we get

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

Ans. Let  be three numbers in A.P.

According to question,

And

Taking , A.P. is (8 – 3), 8, (8 + 3)

5, 8, 11

Taking  A.P. is (8 + 3), 8, (8 – 3)

11, 8, 5

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

Ans. Given:   …..(i)

…..(ii)

And

Now,

Proved.

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

Ans. Given: A.P. 203, 210, 217, ………., 399

Here  and

Now,

= 8729

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

Ans. Given: A.P. which is divisible by 2

2, 4, 6, ………., 100

Here  and

Now,

= 2550

Again A.P. which is divisible by 5

5, 10, 15, ………., 100

Here  and

Now,

= 1050

Again A.P. which is divisible by both 2 and 5

10, 20, 30, ………., 100

Here  and

Now,

= 550

Now, According to question,  = (2550 + 1050) – 550 = 3050

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

Ans. Given: A.P. 13, 17, 21, ………., 97

Here  and

Now,

= 1210

7. If  is a function satisfying  for all  N such that  =3 and  = 120, find the value of

Ans. Here:  and  for all  N ……….(i)

Putting  in eq. (i),

= 9

Putting  in eq. (i),

= 27

Putting  in eq. (i),

= 81

Now,

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. Given:  and

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

Ans. Given:  and

=  =

=

or  which is not possible

Therefore, common ratio is

10. 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.

Ans. Let  be three numbers in G.P., therefore,

..(i)

According to question,   are in A.P.

…..(ii)

Dividing eq. (i) by eq. (ii),

=  =  =

or

Putting  in eq. (i),

Then the required numbers are 8, 16, 32.

Putting  in eq. (i),

Then the required numbers are 32, 16, 8.

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 the number of terms be , number of odd terms be  and  be in G.P.

and

According to question,

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. Given:  and

Also,

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

Ans. Taking

……….(i)

Taking

……….(ii)

From eq. (i) and (ii),

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

Ans. Here , P =  and R =

Now

Proved.

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

Ans. According to question,  ,  and

Now

Putting values of  and  we get

0 = 0 Proved.

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

Ans. Given:  are in A.P.

are in A.P.

are in A.P.

are in A.P. [Adding 1 to each numerator]

are in A.P.

are in A.P.[Dividing each fraction by ]

are in A.P.[Multiplying each fraction by ]

are in A.P.