**NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2**

NCERT Solutions for Class
11 Maths Chapter 9
Sequences and Series Ex 9.2 are the part of NCERT Solutions for Class 11 Maths.
Here you can find the NCERT Solutions for Class 11 Maths Chapter 9 Sequences
and Series Ex 9.2.

**Ex 9.2 Class 11 Maths Question 1.**

Find the sum of odd integers from 1 to 2001.

**Solution:**

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

This series is in 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

**Ex 9.2 Class 11
Maths Question 2.**

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

**Solution:**

**The** multiples of 5 lying between 100 and
1000 are 105, 110, 115, …, 995.

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

This series is in A.P. with first term a = 105, common difference d = 110 – 105
= 5 and last term l = 995.

**Ex 9.2 Class 11
Maths 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 20^{th}term is -112.

**Solution:**

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

**Ex 9.2 Class 11 Maths Question
4.**

How many
terms of the A.P. -6, −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., then

**Ex 9.2 Class 11 Maths Question
5.**

In an A.P.,
if p^{th}term is 1/q and q

^{th}term is 1/p, prove that the sum of first pq terms is ½ (pq + 1), where p ≠ q.

**Solution:**

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

**Ex 9.2 Class 11
Maths 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, S_{n} = 116

**Ex 9.2 Class 11
Maths Question 7.**

Find the sum to n terms of the A.P., whose k^{th}term is 5k + 1.

**Solution:**

Given, a_{k} = 5k + 1

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

**Ex 9.2 Class 11
Maths Question 8.**

If the sum of n terms of an A.P. is (pn + qn^{2}), where p and q are constants, find the common difference.

**Solution:**

Given, S_{n} = pn + qn^{2}, where S_{n} be the sum
of n terms.

**Ex 9.2 Class 11
Maths Question 9.**

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

**Solution:**

Let a_{1}, a_{2} & d_{1}, d_{2} be
the first terms & common differences of the two arithmetic progressions,
respectively. According to the given condition, we have

**Ex 9.2 Class 11
Maths 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 the common difference be d.

According to the question,

**Ex 9.2 Class 11
Maths 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

**Ex 9.2 Class 11
Maths Question 12.**

The ratio of the sums of m and n terms of an A.P. is m^{2 }: n

^{2}. Show that the ratio of m

^{th}and n

^{th}term is (2m – 1) : (2n – 1).

**Solution:**

Let the first term be a & the common difference be d. We have,

**Ex 9.2 Class 11 Maths Question
13.**

If the sum of n terms of an A.P. is 3n^{2}+ 5n and its m

^{th}term is 164, find the value of m.

**Solution:**

Given, S_{n} = 3n^{2} + 5n, where S_{n} be
the sum of n terms.

**Ex 9.2 Class 11
Maths Question 14.**

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

**Solution:**

Let A_{1}, A_{2}, A_{3}, A_{4}, A_{5} be
the numbers between 8 and 26 such that 8, A_{1}, A_{2}, A_{3},
A_{4}, A_{5}, 26 are in A.P. Here a = 8, l = 26, n = 7

**Ex 9.2 Class 11 Maths Question 15.**

If (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, (a^{n }+ b^{n})/(a^{n}^{−}^{1 }+ b^{n}^{−}^{1}) = (a + b)/2

**Ex 9.2 Class 11
Maths 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 7^{th}and (m – 1)

^{th}numbers is 5 : 9. Find the value of m.

**Solution:**

Let A_{1}, A_{2},
………, A_{m} be m numbers which have been inserted between 1 and 31.

Then, the sequence be 1, A_{1}, A_{2}, ……… A_{m}, 31 is
in A.P.

Here, a = 1, 31 is (m + 2)^{th} term,
let d be the common difference.

**Ex 9.2 Class 11
Maths Question 17.**

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

**Solution:**

**The** first, second, third, … instalments are
Rs 100, Rs 105, Rs 110, …

Here, we have an A.P. with a = 100 and d = 5.

We have, a_{n }= a + (n – 1)d

∴ a_{30 }= a + 29d = 100 + 29(5) =
100 + 145 = 245

Hence, the man will pay Rs. 245 in 30^{th} instalment.

**Ex 9.2 Class 11
Maths 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.

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