NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2.

- Sequences and Series Class 11 Ex 9.1
- Sequences and Series Class 11 Ex 9.3
- Sequences and Series Class 11 Ex 9.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Sequences and Series |

Exercise |
Ex 9.2 |

Number of Questions Solved |
18 |

Category |
NCERT Solutions |

## 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 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

**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:**

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

**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 20th term is -112.

**Solution:**

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

**Ex 9.2 Class 11 Maths 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

**Ex 9.2 Class 11 Maths 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

**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:**

We have 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:**

We have S_{n} = pn + qn^{2}, where S„ 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 common difference be d.

According to 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 & common difference be d. Then

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

If the sum of n terms of an A.P. is 3n^{2} + 5n and its mth term is 164, find the value of m.

**Solution:**

We have S_{n} = 3n^{2} + 5n, where Sn 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 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 \(\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n-1 }+{ b }^{ n-1 } } \) is the A.M. between a and b, then find the value of n.

**Solution:**

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

**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 the sequence be 1, A_{1}, A_{2}, ……… A_{m}, 31 Then 31 is (m + 2)^{th} term, a = 1, 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 30th instalment?

**Solution:**

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

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

Hence he 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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