---
title: "NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.3"
url: https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions/exercise-8-3
dateModified: 2026-10-07T15:49:15+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.3

Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions. Every question from Exercise 8.3, with full working and the final answer.

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## Exercise Set 8.3

### Question 1

*2 marks · Very short answer*

Find the $12^{\text{th}}$ term of a GP with common ratio 2, whose $8^{\text{th}}$ term is 192.

**Solution**

1. The $n^{\text{th}}$ term of a GP is given by $t_n = a r^{n-1}$.
2. We are given $t_8 = 192$ and $r = 2$, so $a \times 2^{8-1} = 192 \implies a \times 128 = 192$, which gives $a = \frac{192}{128} = \frac{3}{2}$.
3. The $12^{\text{th}}$ term is $t_{12} = \frac{3}{2} \times 2^{12-1} = \frac{3}{2} \times 2^{11} = 3 \times 2^{10} = 3072$.

**Answer:** 3072

> Common mistake: Confusing the exponent in the formula for the $n^{\text{th}}$ term.

### Question 2

*2 marks · Very short answer*

Find the $10^{\text{th}}$ and $n^{\text{th}}$ terms of the GP: 5, 25, 125, ...

**Solution**

1. The given GP is $5, 25, 125, \dots$, where the first term $a = 5$ and the common ratio $r = \frac{25}{5} = 5$.
2. The $n^{\text{th}}$ term is given by $t_n = a r^{n-1} = 5 \times 5^{n-1} = 5^n$.
3. The $10^{\text{th}}$ term is $t_{10} = 5^{10}$.

**Answer:** $t_{10} = 5^{10}$ and $t_n = 5^n$

> Common mistake: Writing $5^{n-1}$ as $25^{n-1}$.

### Question 3

*3 marks · Short answer*

A sequence is given by the recursive rule $t_1 = 2$, $t_{n+1} = 3t_n - 2$ for $n \geq 1$. Which term of the sequence is 730?

**Solution**

1. The recursive rule is given by $t_1 = 2$ and $t_{n+1} = 3t_n - 2$.
2. Let us find the first few terms: $t_1 = 2$, $t_2 = 3(2) - 2 = 4$, $t_3 = 3(4) - 2 = 10$, $t_4 = 3(10) - 2 = 28$, $t_5 = 3(28) - 2 = 82$, $t_6 = 3(82) - 2 = 244$, $t_7 = 3(244) - 2 = 730$.
3. Thus, 730 is the $7^{\text{th}}$ term of the sequence.

**Answer:** 7th term

> Common mistake: Arithmetic error while evaluating successive terms using the recursive rule.

### Question 4

*3 marks · Short answer*

Which term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the $n^{\text{th}}$ term.

**Solution**

1. The given GP is $2, 6, 18, \dots$, where $a = 2$ and $r = \frac{6}{2} = 3$.
2. The explicit formula for the $n^{\text{th}}$ term is $t_n = 2 \times 3^{n-1}$ and the recursive formula is $t_1 = 2, t_n = 3t_{n-1}$ for $n \geq 2$.
3. To find which term is 4374, substitute $t_n = 4374$ into the explicit formula: $2 \times 3^{n-1} = 4374 \implies 3^{n-1} = 2187$.
4. Since $2187 = 3^7$, we have $n-1 = 7$, which gives $n = 8$. Thus, 4374 is the $8^{\text{th}}$ term.

**Answer:** $8^{\text{th}}$ term, explicit formula $t_n = 2 \times 3^{n-1}$, recursive formula $t_1 = 2, t_n = 3t_{n-1}$

> Common mistake: Incorrectly solving the exponential equation for $n$.

### Question 5

*4 marks · Short answer*

A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way—each time rising to 60% of the previous height.
(i) What height does the ball reach after the $5^{\text{th}}$ bounce?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the $6^{\text{th}}$ time?

**Part (i)**

1. Initial height $a = 80\text{ m}$ and common ratio $r = 0.6$.
2. The height after the $n^{\text{th}}$ bounce is given by $t_n = a r^n$.
3. For the $5^{\text{th}}$ bounce, $t_5 = 80 \times (0.6)^5 = 80 \times 0.07776 = 6.2208\text{ m}$.

Answer (i): $6.2208\text{ m}$

**Part (ii)**

1. The total vertical distance travelled by the time it hits the ground for the $6^{\text{th}}$ time includes the initial fall of $80\text{ m}$ plus twice the sum of the heights of the first 5 bounces.
2. Distance $= 80 + 2 \times (80 \times 0.6 + 80 \times 0.6^2 + 80 \times 0.6^3 + 80 \times 0.6^4 + 80 \times 0.6^5)$.
3. Calculating the heights: $48, 28.8, 17.28, 10.368, 6.2208$, whose sum is $110.6688\text{ m}$.
4. Total distance $= 80 + 2 \times 110.6688 = 80 + 221.3376 = 301.3376\text{ m}$.

Answer (ii): $301.3376\text{ m}$

**Answer:** (i) $6.2208\text{ m}$, (ii) $301.3376\text{ m}$

> Common mistake: Forgetting to multiply the sum of the bounce heights by 2 (since the ball goes up and comes down for every bounce) or forgetting to add the initial drop of 80 m.

### Question 6

*2 marks · Very short answer*

Which term of the sequence $2, 2\sqrt{2}, 4, ...$ is 128?

**Solution**

1. The sequence is $2, 2\sqrt{2}, 4, \dots$, where the first term $a = 2$ and the common ratio $r = \frac{2\sqrt{2}}{2} = \sqrt{2}$.
2. The $n^{\text{th}}$ term is given by $t_n = a r^{n-1} = 2(\sqrt{2})^{n-1} = 2 \times 2^{\frac{n-1}{2}} = 2^{\frac{n+1}{2}}$.
3. Substitute $t_n = 128$: $2^{\frac{n+1}{2}} = 128 = 2^7$, which gives $\frac{n+1}{2} = 7 \implies n + 1 = 14 \implies n = 13$.

**Answer:** $13^{\text{th}}$ term

> Common mistake: Errors in handling fractional exponents with $\sqrt{2}$.

### Question 7

*5 marks · Long answer*

Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on. Look at Fig. 8.12 and try to answer the following questions.
(i) How many red squares are there in Stages 0 to 3?
(ii) Can you predict the number of red squares in Stages 4 and 5?
(iii) Can you find a rule for the number of red squares at the $n^{\text{th}}$ stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.
(iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the $n^{\text{th}}$ stage. What happens to this area as $n$, the number of stages, goes on increasing?

**Part (i)**

1. In Stage 0, there is $1$ main square, so $t_0 = 1$.
2. In Stage 1, there are $8$ red squares.
3. In Stage 2, each of the $8$ squares is divided into $8$ smaller ones, giving $8 \times 8 = 64$ squares.
4. In Stage 3, the number of red squares is $64 \times 8 = 512$.
5. Thus, the number of red squares in Stages 0 to 3 are $1$, $8$, $64$, and $512$ respectively.

Answer (i): Stages 0 to 3 have 1, 8, 64, and 512 red squares respectively.

**Part (ii)**

1. The number of red squares at each stage is multiplied by $8$ to get the next stage.
2. For Stage 4, the number of red squares is $512 \times 8 = 4096$.
3. For Stage 5, the number of red squares is $4096 \times 8 = 32768$.

Answer (ii): Stage 4 has 4,096 red squares and Stage 5 has 32,768 red squares.

**Part (iii)**

1. The sequence of red squares is $1, 8, 64, 512, \dots$, which is a geometric progression with first term $a = 1$ and common ratio $r = 8$.
2. The explicit formula for the $n^{\text{th}}$ stage is $t_n = 8^n$ for $n \ge 0$.
3. The recursive formula is $t_0 = 1$ and $t_n = 8t_{n-1}$ for $n \ge 1$.

Answer (iii): Explicit formula: $t_n = 8^n$; Recursive formula: $t_0 = 1, t_n = 8t_{n-1}$ for $n \ge 1$.

**Part (iv)**

1. In Stage 0, the area is $1$. In Stage 1, one out of nine equal parts is removed, leaving $\frac{8}{9}$ of the area.
2. The areas for Stages 1, 2, and 3 are $\frac{8}{9}$, $(\frac{8}{9})^2 = \frac{64}{81}$, and $(\frac{8}{9})^3 = \frac{512}{729}$ square units respectively.
3. For Stage 4 and Stage 5, the areas are $(\frac{8}{9})^4 = \frac{4096}{6561}$ and $(\frac{8}{9})^5 = \frac{32768}{59049}$ square units.
4. The explicit formula for the area at the $n^{\text{th}}$ stage is $s_n = (\frac{8}{9})^n$ and the recursive formula is $s_0 = 1$, $s_n = \frac{8}{9}s_{n-1}$ for $n \ge 1$.
5. As $n$ goes on increasing, the area of the red region decreases and gets closer and closer to $0$.

Answer (iv): Areas are $\frac{8}{9}, \frac{64}{81}, \frac{512}{729}, \frac{4096}{6561}, \frac{32768}{59049}$; explicit formula $s_n = (\frac{8}{9})^n$; area approaches $0$ as $n$ increases.

**Answer:** The red squares follow $t_n = 8^n$ and the area follows $s_n = (\frac{8}{9})^n$, approaching $0$ as $n$ increases.

> Common mistake: Confusing Stage 0 with Stage 1 when writing the starting values for the geometric progression.

## 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 8.1](https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions/exercise-8-1)
- [Exercise 8.2](https://www.swavid.com/maths/class/9/chapter/predicting-what-comes-next-exploring-sequences-and-progressions/ncert-solutions/exercise-8-2)

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