# Chapter 9 Sequences and Series Ex – 9.2

Question 1.

Find the sum of odd integers from 1 to 2001.

Solution:

We have to find 1 + 3 + 5 + ……….. + 2001

This is an A.P. with first term a = 1, common difference d = 3-1 = 2 and last term l = 2001

∴ l = a + (n-1 )d ⇒ 2001 = 1 + (n -1)2 ⇒ 2001 = 1 + 2n – 2 ⇒ 2n = 2001 + 1

Question 2.

Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Solution:

We have to find 105 +110 +115 + ……..+ 995

This is an A.P. with first term a = 105, common difference d = 110 -105 = 5 and last term 1 = 995

Question 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 20th term is -112.

Solution:

Let a = 2 be the first term and d be the common difference.

Question 4.

How many terms of the A.P. -6, $-\frac { 11 }{ 2 }$, -5, ……. are needed to give the sum – 25 ?

Solution:

Let a be the first term and d be the common difference of the given A.P., we have

Question 5.

In an A.P., if pth term is $\frac { 1 }{ q }$ and qth term is $\frac { 1 }{ p }$ prove that the sum of first pq terms is $\frac { 1 }{ 2 } \left( pq+1 \right)$, where p±q.

Solution:

Let a be the first term & d be the common difference of the A.P., then

Question 6.

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

Solution:

Let a be the first term and d be the common difference.

We have a = 25, d = 22 – 25 = -3, Sn = 116

Question 7.

Find the sum to n terms of the A.P., whose kth term is 5k + 1.

Solution:

We have ak = 5k +1

By substituting the value of k = 1, 2, 3 and 4,

we get

Question 8.

If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.

Solution:

We have Sn = pn + qn2, where S„ be the sum of n terms.

Question 9.

The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.

Solution:

Let a1, a2 & d1 d2 be the first terms & common differences of the two arithmetic progressions respectively. According to the given condition, we have

Question 10.

If the sum of 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.

Solution:

Let the first term be a and common difference be d.

According to question

Question 11.

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

Prove that

Solution:
Let the first term be A & common difference be D. We have
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Question 12.

The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nth term is (2m -1): (2n -1).

Solution:

Let the first term be a & common difference be d. Then

Question 13.

If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.

Solution:

We have Sn = 3n2 + 5n, where Sn be the sum of n terms.

Question 14.

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

Solution:

Let A1, A2, A3, A4, A5 be numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in A.P., Here a = 8, l = 26, n = 7

Question 15.

If $\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n-1 }+{ b }^{ n-1 } }$ is the A.M. between a and b, then find the value of n.

Solution:

We have $\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n-1 }+{ b }^{ n-1 } } =\frac { a+b }{ 2 }$

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

Solution:

Let the sequence be 1, A1, A2, ……… Am, 31 Then 31 is (m + 2)th term, a = 1, let d be the common difference

Question 17.

A man starts repaying a loan as a first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30th installment?

Solution:

Here, we have an A.P. with a = 100 and d = 5
∴ an= a + 29d = 100 + 29(5) = 100 + 145 = 245
Hence he will pay Rs. 245 in 30th instalment.

Question 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.

Solution:

Let there are n sides of a polygon.