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

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## 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
2023-2024.

### 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
- NCERT Solutions And Textual questions Answers Class 11 Mathematics
- Extra NCERT Book questions Answers Class 11 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 9.1 (Available) class 11 Mathematics Chapter 9. Sequences and Series

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- Exercise 9.1 (Available) Class 11 Maths 9. Sequences And Series - Ncert Solutions - Toppers Study
- Class 11 Ncert Solutions
- Solution Chapter 9. Sequences And Series Class 11
- Solutions Class 11
- Chapter 9. Sequences And Series Exercise 9.1 (Available) Class 11

## 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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