Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.3 Solutions

Geometric progression is a sequence in which each term is obtained by multiplying the previous term by a fixed number.
Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.3 connects GP terms, common ratio, recursive formulas, bouncing ball heights and Sierpiński square carpet patterns.

Chapter 8, Predicting What Comes Next: Exploring Sequences and Progressions Class 9, introduces geometric progressions after arithmetic progressions. Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.3 Solutions focus on finding terms of a GP, identifying the position of a term, writing explicit and recursive formulas, solving a bouncing ball GP problem, and studying the Sierpiński square carpet. The textbook explains that a GP is written as a, ar, ar², ar³, …, arⁿ⁻¹, where a is the first term and r is the common ratio.

Key Takeaways

• Geometric Progression: Consecutive terms have a fixed common ratio.
• nth Term of GP: The formula is tn = ar^(n - 1).
• Recursive Formula: A GP can be written as t1 = a, tn = r × tn-1.
• Fractal Pattern: Sierpiński square carpet follows powers of 8 and area powers of 8/9.

Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.3 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.3 Solutions for Geometric Progressions

Exercise 8.3 begins with direct questions on geometric progressions. The key formula is tn = ar^(n - 1), where a is the first term and r is the common ratio.

Q1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.

The 12th term is 3072.

Given:

Common ratio = 2

8th term = 192

Use:

t12 = t8 × r^(12 - 8)

Substitute values:

t12 = 192 × 2^4

t12 = 192 × 16

t12 = 3072

Answer:

The 12th term is 3072.

Ganita Manjari Class 9 Chapter 8 Exercise 8.3: nth Term of GP Class 9

The nth term of a GP helps find faraway terms directly. For the GP 5, 25, 125, …, the first term is 5 and the common ratio is 5.

Q2. Find the 10th and nth terms of the GP: 5, 25, 125, … .

The 10th term is 9765625, and the nth term is 5^n.

Given:

GP = 5, 25, 125, …

First term:

a = 5

Common ratio:

r = 25/5

r = 5

Use:

tn = ar^(n - 1)

Substitute:

tn = 5 × 5^(n - 1)

tn = 5^n

For the 10th term:

t10 = 5^10

t10 = 9765625

Answer:

t10 = 9765625

tn = 5^n

Class 9 Maths Chapter 8 Exercise 8.3 Solutions for Recursive Sequences

A recursive rule gives each term using the previous term. In Question 3, each next term is found by multiplying the earlier term by 3 and subtracting 2.

Q3. A sequence is given by the recursive rule t1 = 2, tn+1 = 3tn - 2 for n ≥ 1. Which term of the sequence is 730?

730 is the 7th term of the sequence.

Given:

t1 = 2

tn+1 = 3tn - 2

Find terms one by one:

t1 = 2

t2 = 3t1 - 2

t2 = 3(2) - 2

t2 = 6 - 2

t2 = 4

t3 = 3t2 - 2

t3 = 3(4) - 2

t3 = 12 - 2

t3 = 10

t4 = 3t3 - 2

t4 = 3(10) - 2

t4 = 30 - 2

t4 = 28

t5 = 3t4 - 2

t5 = 3(28) - 2

t5 = 84 - 2

t5 = 82

t6 = 3t5 - 2

t6 = 3(82) - 2

t6 = 246 - 2

t6 = 244

t7 = 3t6 - 2

t7 = 3(244) - 2

t7 = 732 - 2

t7 = 730

Answer:

730 is the 7th term.

Class 9 Maths Ganita Manjari Chapter 8 Solutions for GP Formula

A GP can be written using an explicit formula and a recursive formula. The explicit formula gives the nth term directly, while the recursive formula gives the next term from the previous term.

Q4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.

4374 is the 8th term of the GP.

Given:

GP = 2, 6, 18, …

First term:

a = 2

Common ratio:

r = 6/2

r = 3

Use:

tn = ar^(n - 1)

So:

tn = 2 × 3^(n - 1)

Now set:

2 × 3^(n - 1) = 4374

Divide by 2:

3^(n - 1) = 4374/2

3^(n - 1) = 2187

Since:

2187 = 3^7

So:

3^(n - 1) = 3^7

n - 1 = 7

n = 8

Explicit formula:

tn = 2 × 3^(n - 1)

Recursive formula:

t1 = 2

tn = 3tn-1 for n ≥ 2

Answer:

4374 is the 8th term.

Explicit formula: tn = 2 × 3^(n - 1)

Recursive formula: t1 = 2, tn = 3tn-1 for n ≥ 2

Predicting What Comes Next Exploring Sequences and Progressions Class 9: Bouncing Ball GP Problem

The bouncing ball problem forms a GP because each bounce reaches 60% of the previous height. The common ratio is 0.6.

Q5. 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.

Q5(i). What height does the ball reach after the 5th bounce?

The ball reaches 6.2208 m after the 5th bounce.

Given:

Initial height = 80 m

Bounce ratio = 60%

Bounce ratio = 0.6

Height after 1st bounce:

h1 = 80 × 0.6

h1 = 48 m

Height after 2nd bounce:

h2 = 48 × 0.6

h2 = 28.8 m

Height after 3rd bounce:

h3 = 28.8 × 0.6

h3 = 17.28 m

Height after 4th bounce:

h4 = 17.28 × 0.6

h4 = 10.368 m

Height after 5th bounce:

h5 = 10.368 × 0.6

h5 = 6.2208 m

Answer:

The ball reaches 6.2208 m after the 5th bounce.

Q5(ii). What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?

The total vertical distance is 301.3376 m.

The ball first falls 80 m.

Then it rises and falls after each of the first five bounces.

Bounce heights:

h1 = 48 m

h2 = 28.8 m

h3 = 17.28 m

h4 = 10.368 m

h5 = 6.2208 m

Total distance:

Distance = initial fall + 2(h1 + h2 + h3 + h4 + h5)

Distance = 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208)

Distance = 80 + 2(110.6688)

Distance = 80 + 221.3376

Distance = 301.3376 m

Answer:

The total vertical distance travelled is 301.3376 m.

Class 9 Maths Sequences Solutions: Finding the Term in a Radical GP

A geometric progression can also have an irrational common ratio. In the sequence 2, 2√2, 4, …, each term is multiplied by √2.

Q6. Which term of the sequence 2, 2√2, 4, … is 128?

128 is the 13th term of the sequence.

Given:

Sequence = 2, 2√2, 4, …

First term:

a = 2

Common ratio:

r = 2√2 / 2

r = √2

Use:

tn = ar^(n - 1)

So:

tn = 2(√2)^(n - 1)

Now set:

2(√2)^(n - 1) = 128

Write powers of 2:

2 × (2^(1/2))^(n - 1) = 2^7

2 × 2^((n - 1)/2) = 2^7

2^(1 + (n - 1)/2) = 2^7

Equate powers:

1 + (n - 1)/2 = 7

(n - 1)/2 = 6

n - 1 = 12

n = 13

Answer:

128 is the 13th term.

Class 9 Maths Chapter 8 Exercise 8.3 Solutions for Sierpiński Square Carpet

The Sierpiński square carpet is a fractal pattern. At each stage, every retained square gives 8 smaller red squares in the next stage. The textbook shows Stages 0 to 3 of this pattern in Fig. 8.12.

Q7. Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet.

Q7(i). How many red squares are there in Stages 0 to 3?

The numbers of red squares are 1, 8, 64 and 512.

Stage 0:

Number of red squares = 1

Stage 1:

Number of red squares = 8

Stage 2:

Number of red squares = 8 × 8

Number of red squares = 64

Stage 3:

Number of red squares = 64 × 8

Number of red squares = 512

Answer:

Stage 0 = 1

Stage 1 = 8

Stage 2 = 64

Stage 3 = 512

Q7(ii). Can you predict the number of red squares in Stages 4 and 5?

The numbers of red squares in Stages 4 and 5 are 4096 and 32768.

Stage 4:

Number of red squares = 512 × 8

Number of red squares = 4096

Stage 5:

Number of red squares = 4096 × 8

Number of red squares = 32768

Answer:

Stage 4 = 4096

Stage 5 = 32768

Q7(iii). Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula.

The number of red squares at Stage n is 8^n.

The pattern is:

1, 8, 64, 512, …

This can be written as:

8^0, 8^1, 8^2, 8^3, …

Explicit formula:

tn = 8^n

Recursive formula:

t0 = 1

tn = 8tn-1 for n ≥ 1

Answer:

Explicit formula: tn = 8^n

Recursive formula: t0 = 1, tn = 8tn-1 for n ≥ 1

Q7(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 in Stages 4 and 5? Find the explicit and recursive formulas for the area of the red region at the nth stage. What happens as n increases?

The red area at Stage n is (8/9)^n square units.

At each stage, the square is divided into 9 equal smaller squares.

The centre square is removed.

So the remaining red area becomes:

8/9 of the previous red area

Stage 0:

Area = 1

Stage 1:

Area = 8/9

Stage 2:

Area = (8/9)^2

Area = 64/81

Stage 3:

Area = (8/9)^3

Area = 512/729

Stage 4:

Area = (8/9)^4

Area = 4096/6561

Stage 5:

Area = (8/9)^5

Area = 32768/59049

Explicit formula:

sn = (8/9)^n

Recursive formula:

s0 = 1

sn = (8/9)sn-1 for n ≥ 1

As n increases, the red area keeps decreasing and gets closer to 0.

Answer:

Stage 1 = 8/9

Stage 2 = 64/81

Stage 3 = 512/729

Stage 4 = 4096/6561

Stage 5 = 32768/59049

Explicit formula: sn = (8/9)^n

Recursive formula: s0 = 1, sn = (8/9)sn-1 for n ≥ 1

Geometric Progression Class 9: Concepts Used in Exercise 8.3

Exercise 8.3 uses geometric progressions, recursive formulas and fractal growth. The main idea is repeated multiplication by the same number.

Geometric Progression Class 9

A geometric progression is a sequence in which each term is found by multiplying the previous term by a fixed number.

Copy-friendly example:

2, 6, 18, 54, …

Here:

Common ratio = 3

Common Ratio Class 9

The common ratio is found by dividing a term by the previous term.

Copy-friendly formula:

r = second term / first term

Example:

For 2, 6, 18, …

r = 6/2

r = 3

nth Term of GP Class 9

The nth term of a GP gives the term at position n.

Copy-friendly formula:

tn = ar^(n - 1)

Here:

a = first term

r = common ratio

n = term position

Recursive Formula for GP

A recursive formula gives the next term using the previous term.

Copy-friendly formula:

t1 = a

tn = rtn-1 for n ≥ 2

Example:

For 2, 6, 18, …

t1 = 2

tn = 3tn-1 for n ≥ 2

Sierpiński Square Carpet Class 9

The Sierpiński square carpet creates a GP because each retained square gives 8 retained squares in the next stage.

Copy-friendly results:

Number of red squares at Stage n = 8^n

Area of red region at Stage n = (8/9)^n

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.3 Solutions

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8