Ncert solution for class 11 mathematics chapter - 9 SEQUENCES AND SERIES

Ncert solution for class 11 mathematics chapter - 9 SEQUENCES AND SERIES

Exercise 9.1

1. Ncert Solution Class 11 question 1 to 6 
2. Ncert Solution for class 11 question 7 to 10
3. Ncert  Solutions for exercise 9.1 question 11 to 13 
4. Ncert solution question 14

Exercise 9.2

1.  Find the sum of odd integers from 1 to 2001. 
2.  Find the sum of all natural numbers lying between 100 and 1000, which aremultiples of 5.
3.  In an A.P., the first term is 2 and the sum of the first five terms is one-fourth ofthe next five terms. Show that 20th term is –112. 
4.  How many terms of the A.P. – 6, 112− , – 5, … are needed to give the sum –25?
5.  In an A.P., if pth term is 1qand qth term is 1p, prove that the sum of first pqterms is 12(pq +1), where p ≠ q. 
6.  If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find thelast term. 
7.  Find the sum to n terms of the A.P., whose kth term is 5k + 1. 
8.  If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants,find the common difference. 
9.  The sums of n terms of two arithmetic progressions are in the ratio5n + 4 : 9n + 6. Find the ratio of their 18th terms. 
10.  If the sum of first p terms of an A.P. is equal to the sum of the first q terms, thenfind the sum of the first (p + q) terms. 
11.  Sum of the first p, q and r terms of an A.P are. a, b and c, respectively.Prove that a (q r) b (r p) c ( p q) 0p q r− + − + − =12. The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratioof mth and nth term is (2m – 1) : (2n – 1). 
12.  If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the valueof m. 
13.  Insert five numbers between 8 and 26 such that the resulting sequence is an A.P. 
14.  If 1 1n nn na ba − b −++is the A.M. between a and b, then find the value of n. 
15. Between 1 and 31, m numbers have been inserted in such a way that the resultingsequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find thevalue of m. 
16.  A man starts repaying a loan as first instalment of Rs. 100. If he increases theinstalment by Rs 5 every month, what amount he will pay in the 30th instalment? 
17. 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. 

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.

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.

Answer

The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°.
It is known that the sum of all angles of a polygon with n sides is 180°

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 30th installment?

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 30th installment?

Answer

The first installment of the loan is Rs 100.
The second installment of the loan is Rs 105 and so on.
The amount that the man repays every month forms an A.P.
The A.P. is 100, 105, 110, …
First term, a = 100
Common difference, d = 5
A30 = a + (30 – 1)d
= 100 + (29) (5)
= 100 + 145
= 245
Thus, the amount to be paid in the 30th installment is Rs 245.

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 7th and (m – 1)th numbers is 5:9. Find the value of m.

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 7th and (m – 1)th numbers is 5:9. Find the value of m.

Answer

Let A1, A2, … Am be m numbers such that 1, A1, A2, … Am, 31 is an A.P.
Here, a = 1, b = 31, n = m + 2
∴ 31 = 1 + (m + 2 – 1) (d)
⇒ 30 = (m + 1) d

d = 30/(m+1)                ...(1)
A1 = a + d
A2 = a + 2d
A3 = a + 3d …
∴ A7 = a + 7d
Am–1 = a + (m – 1) d
According to the given condition,