---
title: "NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.1"
url: https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions/exercise-8-1
dateModified: 2026-10-07T15:49:15+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.1

Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions. Every question from Exercise 8.1, with full working and the final answer.

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## Exercise Set 8.1

### Question 1

*3 marks · Short answer*

Find the first five terms of the sequence in which the $n^{\text{th}}$ term is given by (i) $t_n = 3n - 4$, (ii) $t_n = 2 - 5n$, and (iii) $t_n = n^2 - 2n + 3$ for $n \geq 1$.

**Part (i)**

1. For $t_n = 3n - 4$, substitute $n = 1, 2, 3, 4, 5$.
2. $t_1 = 3(1) - 4 = -1$, $t_2 = 3(2) - 4 = 2$, $t_3 = 3(3) - 4 = 5$.
3. $t_4 = 3(4) - 4 = 8$, $t_5 = 3(5) - 4 = 11$.

Answer (i): -1, 2, 5, 8, 11

**Part (ii)**

1. For $t_n = 2 - 5n$, substitute $n = 1, 2, 3, 4, 5$.
2. $t_1 = 2 - 5(1) = -3$, $t_2 = 2 - 5(2) = -8$, $t_3 = 2 - 5(3) = -13$.
3. $t_4 = 2 - 5(4) = -18$, $t_5 = 2 - 5(5) = -23$.

Answer (ii): -3, -8, -13, -18, -23

**Part (iii)**

1. For $t_n = n^2 - 2n + 3$, substitute $n = 1, 2, 3, 4, 5$.
2. $t_1 = 1^2 - 2(1) + 3 = 2$, $t_2 = 2^2 - 2(2) + 3 = 3$, $t_3 = 3^2 - 2(3) + 3 = 6$.
3. $t_4 = 4^2 - 2(4) + 3 = 11$, $t_5 = 5^2 - 2(5) + 3 = 18$.

Answer (iii): 2, 3, 6, 11, 18

**Answer:** (i) -1, 2, 5, 8, 11; (ii) -3, -8, -13, -18, -23; (iii) 2, 3, 6, 11, 18

> Common mistake: Arithmetic errors when substituting negative values or powers.

### Question 2

*2 marks · Very short answer*

Find the $10^{\text{th}}$ and $15^{\text{th}}$ terms of the sequence $t_n = 5n - 3$ for $n \geq 1$.

**Solution**

1. The given explicit formula is $t_n = 5n - 3$.
2. Substitute $n = 10$ to get $t_{10} = 5(10) - 3 = 47$.
3. Substitute $n = 15$ to get $t_{15} = 5(15) - 3 = 72$.

**Answer:** The $10^{\text{th}}$ term is $47$ and the $15^{\text{th}}$ term is $72$.

> Common mistake: Multiplying incorrectly or making errors in subtraction.

### Question 3

*3 marks · Short answer*

Determine whether 97 and 172 are terms of the sequence $t_n = 5n - 3$ for $n \geq 1$.

**Solution**

1. Equate the given explicit formula $t_n = 5n - 3$ to 97 to check if 97 is a term.
2. $5n - 3 = 97 \implies 5n = 100 \implies n = 20$. Since 20 is a natural number, 97 is a term.
3. Equate $5n - 3$ to 172 to check if 172 is a term.
4. $5n - 3 = 172 \implies 5n = 175 \implies n = 35$. Since 35 is a natural number, 172 is a term.

**Answer:** Both 97 and 172 are terms of the sequence (specifically the $20^{\text{th}}$ and $35^{\text{th}}$ terms respectively).

> Common mistake: Assuming a number is not a term without fully solving for $n$ as a natural number.

### Question 4

*2 marks · Very short answer*

Which term of the sequence $t_n = 5n - 3$ for $n \geq 1$ is 607?

**Solution**

1. Set $t_n = 607$ using the formula $t_n = 5n - 3$.
2. $5n - 3 = 607 \implies 5n = 610 \implies n = 122$.

**Answer:** 607 is the $122^{\text{nd}}$ term of the sequence.

> Common mistake: Calculation error while dividing 610 by 5.

### Question 5

*3 marks · Short answer*

A sequence is given by the recursive rule $t_1 = -5$, $t_{n+1} = t_n + 3$ for $n \geq 1$. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?

**Solution**

1. Given $t_1 = -5$ and $t_{n+1} = t_n + 3$.
2. $t_2 = t_1 + 3 = -5 + 3 = -2$, $t_3 = t_2 + 3 = -2 + 3 = 1$, $t_4 = t_3 + 3 = 1 + 3 = 4$, $t_5 = t_4 + 3 = 4 + 3 = 7$.
3. To check if 52 is a term, we observe that the sequence is an AP with $a = -5$ and $d = 3$, so its explicit rule is $t_n = a + (n-1)d = -5 + (n-1)3 = 3n - 8$.
4. Set $3n - 8 = 52 \implies 3n = 60 \implies n = 20$. Since 20 is a natural number, 52 is a term.

**Answer:** The first five terms are $-5, -2, 1, 4, 7$. Yes, 52 is a term, and it is the $20^{\text{th}}$ term.

> Common mistake: Mistaking the recursive formula to find the position directly without finding the explicit formula or stepping through.

### Question 6

*3 marks · Short answer*

Let $T_1 = 1$, $T_2 = 2$, $T_3 = 4$, and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \geq 4$. Find $T_4$, $T_5$, $T_6$, $T_7$, and $T_8$.

**Solution**

1. Given $T_1 = 1$, $T_2 = 2$, $T_3 = 4$, and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \geq 4$.
2. $T_4 = T_3 + T_2 + T_1 = 4 + 2 + 1 = 7$.
3. $T_5 = T_4 + T_3 + T_2 = 7 + 4 + 2 = 13$, $T_6 = T_5 + T_4 + T_3 = 13 + 7 + 4 = 24$.
4. $T_7 = T_6 + T_5 + T_4 = 24 + 13 + 7 = 44$, $T_8 = T_7 + T_6 + T_5 = 44 + 24 + 13 = 81$.

**Answer:** $T_4 = 7$, $T_5 = 13$, $T_6 = 24$, $T_7 = 44$, $T_8 = 81$.

> Common mistake: Adding wrong previous terms due to indexing confusion.

## Related pages

- [All Chapter 8 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions)
- [Exercise 8.2](https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions/exercise-8-2)
- [Exercise 8.3](https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions/exercise-8-3)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student