NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 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 9.4.

- Sequences and Series Class 11 Ex 9.1
- Sequences and Series Class 11 Ex 9.2
- Sequences and Series Class 11 Ex 9.3

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Sequences and Series |

Exercise |
Ex 9.4 |

Number of Questions Solved |
10 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4

**Find the sunt to n terms of each of the series in Exercises 1 to 7.**

**Ex 9.4 Class 11 Maths Question 1.**

1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ………

**Solution:**

In the given series, there is a sum of multiple of corresponding terms of two A.P’s. The two A.P’s are

**Ex 9.4 Class 11 Maths Question 2.**

1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ……

**Solution:**

In the given series, there is a sum of multiple of corresponding terms of two A.P’s. The three A.P’s are

**Ex 9.4 Class 11 Maths Question 3.**

3 x 1^{2} + 5 x 2^{2} + 7 x 3^{2} + …..

**Solution:**

In the given series there is sum of multiple of corresponding terms of two A.P’s. The two A.P’s are

(i) 3, 5, 7, …………… and

(ii) 1^{2}, 2^{2}, 3^{2}, ………………….

Now the n^{th} term of sum is an = (nth term of the sequence formed by first A.P.) x (n^{th} term of the sequence formed by second A.P.) = (2 n + 1) x n^{2} = 2n^{3} + n^{2} Hence, the sum to n terms is,

**Ex 9.4 Class 11 Maths Question 4.**

\(\frac { 1 }{ 1\times 2 } +\frac { 1 }{ 2\times 3 } +\frac { 1 }{ 3\times 4 } +\) …….

**Solution:**

In the given series there is sum of multiple of corresponding terms of two A.P’s. The two A.P’s are

**Ex 9.4 Class 11 Maths Question 5.**

5^{2} + 6^{2} + 7^{2} + ………….. + 20^{2}

**Solution:**

The given series can be written in the following way

**Ex 9.4 Class 11 Maths Question 6.**

3 x 8 + 6 x 11 + 9 x 25 + ………….

**Solution:**

In the given series, there is sum of multiple of corresponding terms of two A.P/s. The two A.P/s are

(i) 3, 6, 9, ………….. and

(ii) 8, 11, 14, ……………….

Now the nth term of sum is an = (n^{th} term of the sequence formed by first A.P.) x (n^{th} term of the sequence formed by second A.P.)

**Ex 9.4 Class 11 Maths Question 7.**

1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + ………….

**Solution:**

In the given series

a_{n} = 1^{2} + 2^{2} + …………….. + n^{2}

Find the sum to n terms of the series in Exercises 8 to 10 whose n^{th} terms is given by

**Ex 9.4 Class 11 Maths Question 8.**

n(n + 1)(n + 4)

**Solution:**

We have

**Ex 9.4 Class 11 Maths Question 9.**

n^{2} + 2^{n}

**Solution:**

We have a_{n} = n^{2} + 2^{n}

Hence, the sum to n terms is,

**Ex 9.4 Class 11 Maths Question 10.**

(2n – 1)^{2}

**Solution:**

We have

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