NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 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 9.1.

- Sequences and Series Class 11 Ex 9.2
- 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 9.1 |

Number of Questions Solved |
14 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

**Question 1.**

a_{n} = n(n + 2)

**Solution:**

We haven a_{n} = n(n + 2)

subtituting n = 1, 2, 3, 4, 5, we get

a_{1} = 9(1 + 2) = 1 x 3 = 3

a_{2} = 2(2 + ) = 2 x 4 = 8

a_{3} = 3(3 + 2) = 3 x 5 = 15

a_{4} = 4(4 + 2) = 4 x 6 = 24

a_{5} = 5(5 + 2) = 5 x 7 = 35

∴ The first five terms are 3, 8, 15, 24, 35.

**Question 2.**

a_{n} =

**Solution:**

**Question 3.**

a_{n} = 2^{n}

**Solution:**

**Question 4.**

a_{n} =

**Solution:**

**Question 5.**

a_{n} = (- 1)^{n-1} 5^{n+1}

**Solution:**

We have, a_{n} = (- 1)^{n-1} 5^{n+1}

Substituting n = 1, 2, 3, 4, 5, we get

a_{1} =(-1)^{1-1} 5^{1+1} = (-1)^{°} 5^{2} = 25

a_{2} =(-1)^{2-1} 5^{2+1} = (-1)^{1} 5^{3} = 125

a_{3} =(-1)^{3-1} 5^{3+1} = (-1)^{2} 5^{4} = 625

a_{4} =(-1)^{4-1} 5^{4+1} = (-1)^{3} 5^{5} = -3125

a_{5} =(-1)^{5-1} 5^{5+1} = (-1)^{4} 5^{6} = 15625

∴ The first five terms are 25, – 125, 625, -3125, 15625.

**Question 6.**

a_{n} = n

**Solution:**

Substituting n = 1, 2, 3, 4, 5, we get

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

**Question 7.**

a_{n} = 4n – 3; a_{17}, a_{24}

**Solution:**

We have a_{n} = 4n – 3

**Question 8.**

a_{n} = ; a_{7}

**Solution:**

We have, a_{n} = ; a_{7}

**Question 9.**

a_{n} = (-1)^{n – 1} n^{3}; a_{9}

**Solution:**

We have, a_{n} = (-1)^{n – 1} n^{3}

**Question 10.**

a_{n} = ; a _{20}

**Solution:**

We have, a_{n} =

**Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:**

**Question 11.**

a_{1} = 3, a_{n} = 3a_{n-1}+2 for all n>1

**Solution:**

We have given a_{1} = 3, a_{n} = 3a_{n-1}+2

⇒ a_{1} = 3, a_{2} = 3a_{1} + 2 = 3.3 + 2 = 9 + 2 = 11,

a_{3} = 3a_{2} + 2 = 3.11 + 2 = 33 + 2 = 35,

a_{4} = 3a_{3} + 2 = 3.35 + 2 = 105 + 2 = 107,

a_{5} = 3a_{4} + 2 = 3.107 + 2 = 321 + 2 = 323,

Hence, the first five terms of the sequence are 3, 11, 35, 107, 323.

The corresponding series is 3 + 11 + 35 + 107 + 323 + ………..

**Question 12.**

a_{1} = -1, a_{n} = , n ≥ 2

**Solution:**

We have given

Hence the first five terms of the given sequence are – 1, -1/2, -1/6, -1/24, -1/120.

The corresponding series is

**Question 13.**

a_{1} = a_{2} = 2, a_{n} = a_{n-1} – 1, n >2

**Solution:**

We have given a_{1} = a_{2} = 2, a_{n} = a_{n-1} – 1, n >2

a_{1} = 2, a_{2} = 2, a_{3}= a_{2} – 1 = 2 – 1 = 1,

a_{4} = a_{3} – 1 = 1 – 1 = 0 and a_{5} = a_{4} – 1 = 0 – 1 = -1

Hence the first five terms of the sequence are 2, 2, 1, 0, -1

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

**Question 14.**

Find Fibonacci sequence is defined by 1 = a_{1} = a_{2} and a_{n} = a_{n-1} + a_{n-2}, n > 2

Find , for n = 1, 2, 3, 4, 5

**Solution:**

We have,

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