Our ncert solutions for Exercise 9.1 (Available) Class 11 maths 9. Sequences and Series - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 9.1 (Available) Class 11 maths 9. Sequences and Series - ncert solutions - Toppers Study

Topper Study classes prepares ncert solutions on practical base problems and comes out with the best result that helps the students and teachers as well as tutors and so many ecademic coaching classes that they need in practical life. Our ncert solutions for Exercise 9.1 (Available) Class 11 maths 9. Sequences and Series - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

## Exercise 9.1 (Available) Class 11 maths 9. Sequences and Series - ncert solutions - Toppers Study

CBSE board students who preparing for **class 11 ncert solutions maths and Mathematics** solved exercise **chapter 9. Sequences and Series** available and this helps in upcoming exams
2022-2023.

### You can Find Mathematics solution Class 11 Chapter 9. Sequences and Series

- All Chapter review quick revision notes for chapter 9. Sequences and Series Class 11
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### Exercise 9.1 (Available) class 11 Mathematics Chapter 9. Sequences and Series

## 9. Sequences and Series

### | Exercise 9.1 (Available) |

## Exercise 9.1 (Available) Class 11 maths 9. Sequences and Series - ncert solutions - Toppers Study

**Exercise 9.1**

Write the first five terms of each of the sequences in Exercises 1 to 6 whose *n*^{th} terms are:

Q1. Write the first five terms of the sequences whose n^{th} term is *a _{n} = n(n+2).*

*Solution:**Q1.*

*is a _{n} = n(n+2)*

*Putting n = 1, 2, 3, 4 and 5 we obtain*

*a _{1} = 1(1 + 2) = 1× 3 = 3 *

*a _{2} = 2(2 + 2) = 2× 4 = 8 *

*a _{3} = 3(3 + 2) = 3× 5 = 15 *

*a _{4} = 4(4 + 2) = 4× 6 = 24 *

*a _{5} = 5(5 + 2) = 5× 7 = 35 *

Therefore, the five terms are 3, 8, 15, 24 and 35

*Q5. *Write the first five terms of the sequences whose n^{th} term is *a _{n} = (-1)^{n-1}5^{n+1}*

*Solution: Q5. *

*a _{n} = (-1)^{n-1}5^{n+1}*

Putting n = 1, 2, 3, 4 and 5 we obtain

*a _{1} = (-1)^{1-1}5^{1+1} = (1)(5)^{2} = 25*

*a _{2} = (-1)^{2-1}5^{2+1} = (-1) (5)^{3} = - 125*

*a _{3} = (-1)^{3-1}5^{3+1} = (1) (5)^{4} = 625*

*a _{4} = (-1)^{4-1}5^{4+1} = (-1) (5)^{5} = - 3125*

*a _{5} = (-1)^{5-1}5^{5+1} = (1) (5)^{6} =156 25*

Thus the five terms are 25, -125, 625, - 3125 and 15625

##### Other Pages of this Chapter: 9. Sequences and Series

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