Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises

Sequences are ordered lists of numbers, and each number in a sequence is called a term.
Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises connect arithmetic progressions, geometric progressions, recursive rules and sums of consecutive natural numbers.

Chapter 8, Predicting What Comes Next Exploring Sequences and Progressions Class 9, brings together explicit formulas, recursive formulas, arithmetic progressions and geometric progressions. Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises include 15 questions on AP term-finding, GP formulas, multiples, consecutive natural numbers, bacteria growth and recursive sequences. These Class 9 Maths Chapter 8 End of Chapter Exercise Solutions follow the textbook order and use copy-friendly equations. The textbook defines an arithmetic progression as a sequence with a constant difference and a geometric progression as a sequence with a constant ratio.

Key Takeaways

• AP Formula: The nth term of an AP is tn = a + (n - 1)d.
• GP Formula: The nth term of a GP is tn = ar^(n - 1).
• Consecutive Sum: 1 + 2 + 3 + ... + n = n(n + 1)/2.
• Recursive Sequence: A term can be defined using one or more previous terms.

Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises Structure 2026

Exercise Type	Topic	Question Count
AP Questions	nth term, multiples and sums	7
GP Questions	terms, ratio and proof	5
Recursive Questions	bacteria growth and special sequences	3

Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises for AP Terms

The first four questions use the arithmetic progression formula. Students find the first term, common difference, term position and number of multiples.

Q1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

The 31st term is 178.

Given:

t11 = 38

t16 = 73

Use:

tn = a + (n - 1)d

For the 11th term:

a + 10d = 38

For the 16th term:

a + 15d = 73

Subtract:

(a + 15d) - (a + 10d) = 73 - 38

5d = 35

d = 7

Now substitute in:

a + 10d = 38

a + 10(7) = 38

a + 70 = 38

a = -32

Find the 31st term:

t31 = a + 30d

t31 = -32 + 30(7)

t31 = -32 + 210

t31 = 178

Answer:

The 31st term is 178.

Q2. Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

The AP is 4, 10, 16, 22, 28, ...

Given:

t3 = 16

t7 - t5 = 12

Use:

tn = a + (n - 1)d

For the third term:

a + 2d = 16

Now:

t7 = a + 6d

t5 = a + 4d

So:

t7 - t5 = 12

(a + 6d) - (a + 4d) = 12

2d = 12

d = 6

Substitute in:

a + 2d = 16

a + 2(6) = 16

a + 12 = 16

a = 4

So the AP is:

4, 10, 16, 22, 28, ...

Answer:

The required AP is 4, 10, 16, 22, 28, ...

Ganita Manjari Class 9 Chapter 8 End of Chapter Exercises: Multiples in AP

Multiples of a fixed number form an arithmetic progression. The common difference is the same as the number whose multiples are being counted.

Q3. How many three-digit numbers are divisible by 7?

There are 128 three-digit numbers divisible by 7.

The smallest three-digit multiple of 7 is:

105

The largest three-digit multiple of 7 is:

994

The AP is:

105, 112, 119, ..., 994

Here:

a = 105

d = 7

tn = 994

Use:

tn = a + (n - 1)d

Substitute:

994 = 105 + (n - 1)7

994 - 105 = 7(n - 1)

889 = 7(n - 1)

127 = n - 1

n = 128

Answer:

There are 128 three-digit numbers divisible by 7.

Q4. How many multiples of 4 lie between 10 and 250?

There are 60 multiples of 4 between 10 and 250.

The smallest multiple of 4 greater than 10 is:

12

The largest multiple of 4 less than 250 is:

248

The AP is:

12, 16, 20, ..., 248

Here:

a = 12

d = 4

tn = 248

Use:

tn = a + (n - 1)d

Substitute:

248 = 12 + (n - 1)4

248 - 12 = 4(n - 1)

236 = 4(n - 1)

59 = n - 1

n = 60

Answer:

There are 60 multiples of 4 between 10 and 250.

Class 9 Maths Chapter 8 End of Chapter Exercise Solutions for GP Conditions

A geometric progression uses a common ratio. In these questions, unknown terms are found using GP relations.

Q5. Find a GP for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.

Two possible GPs are -4/3, -8/3, -16/3, ... and 4, -8, 16, ...

Let the first term be a.

Let the common ratio be r.

First two terms:

a and ar

Given:

a + ar = -4

So:

a(1 + r) = -4

Also:

fifth term = 4 × third term

ar^4 = 4ar^2

Since a is non-zero:

r^4 = 4r^2

r^2 = 4

So:

r = 2 or r = -2

Case 1:

r = 2

a(1 + 2) = -4

3a = -4

a = -4/3

GP:

-4/3, -8/3, -16/3, ...

Case 2:

r = -2

a(1 - 2) = -4

-a = -4

a = 4

GP:

4, -8, 16, -32, ...

Answer:

Two possible GPs are -4/3, -8/3, -16/3, ... and 4, -8, 16, -32, ...

Class 9 Maths Sequences Solutions for Sum of Consecutive Natural Numbers

A number can be written as a sum of consecutive natural numbers when a suitable starting number and number of terms satisfy the sum formula.

Q6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

The possible ways are 100, 18 + 19 + 20 + 21 + 22, and 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16.

Let there be k consecutive natural numbers starting from m.

Then:

100 = m + (m + 1) + (m + 2) + ... + (m + k - 1)

Use:

Sum = k/2 × [2m + k - 1]

So:

100 = k/2 × [2m + k - 1]

200 = k[2m + k - 1]

Check positive integer values of k that divide 200.

Valid cases:

For k = 1:

100 = 100

For k = 5:

2m + 4 = 40

2m = 36

m = 18

So:

100 = 18 + 19 + 20 + 21 + 22

For k = 8:

2m + 7 = 25

2m = 18

m = 9

So:

100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

Answer:

100 = 100

100 = 18 + 19 + 20 + 21 + 22

100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

Predicting What Comes Next Exploring Sequences and Progressions Class 9: Growth Sequences

Bacteria growth is modelled by a geometric progression because the quantity doubles every hour.

Q7. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?

At the end of the 2nd hour there are 120 bacteria, at the end of the 4th hour there are 480 bacteria, and at the end of the nth hour there are 30 × 2^n bacteria.

Given:

Original number of bacteria = 30

Growth factor per hour = 2

At the end of 1st hour:

30 × 2 = 60

At the end of 2nd hour:

30 × 2^2 = 120

At the end of 4th hour:

30 × 2^4 = 30 × 16

30 × 2^4 = 480

At the end of nth hour:

30 × 2^n

Answer:

2nd hour = 120 bacteria

4th hour = 480 bacteria

nth hour = 30 × 2^n bacteria

Class 9 Maths Ganita Manjari Chapter 8 Solutions for AP Equations

When sums of two AP terms are given, write each term using a and d. Then solve the two linear equations.

Q8. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

The first three terms are -13, -8 and -3.

Use:

tn = a + (n - 1)d

Given:

t4 + t8 = 24

(a + 3d) + (a + 7d) = 24

2a + 10d = 24

a + 5d = 12

Also:

t6 + t10 = 44

(a + 5d) + (a + 9d) = 44

2a + 14d = 44

a + 7d = 22

Subtract:

(a + 7d) - (a + 5d) = 22 - 12

2d = 10

d = 5

Substitute in:

a + 5d = 12

a + 5(5) = 12

a + 25 = 12

a = -13

First three terms:

a = -13

a + d = -13 + 5 = -8

a + 2d = -13 + 10 = -3

Answer:

The first three terms are -13, -8 and -3.

Q9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

The smallest value of n is 45.

Use:

Sum of first n natural numbers = n(n + 1)/2

We need:

n(n + 1)/2 > 1000

Check n = 44:

44 × 45 / 2 = 990

990 is less than 1000.

Check n = 45:

45 × 46 / 2 = 1035

1035 is greater than 1000.

Answer:

The smallest value of n is 45.

Class 9 Maths Chapter 8 End of Chapter Exercise Solutions for GP Formula

A GP term is found using the common ratio. The recursive formula is written by multiplying the previous term by the common ratio.

Q10. Which term of the GP: 2, 8, 32, ... is 131072? Write the explicit formula as well as the recursive formula for the nth term.

131072 is the 9th term.

Given:

GP = 2, 8, 32, ...

First term:

a = 2

Common ratio:

r = 8/2

r = 4

Explicit formula:

tn = ar^(n - 1)

tn = 2 × 4^(n - 1)

Now set:

2 × 4^(n - 1) = 131072

Write 131072 as a power of 2:

131072 = 2^17

Also:

2 × 4^(n - 1) = 2 × (2^2)^(n - 1)

2 × 4^(n - 1) = 2 × 2^(2n - 2)

2 × 4^(n - 1) = 2^(2n - 1)

So:

2^(2n - 1) = 2^17

2n - 1 = 17

2n = 18

n = 9

Recursive formula:

t1 = 2

tn = 4tn-1 for n ≥ 2

Answer:

131072 is the 9th term.

Explicit formula: tn = 2 × 4^(n - 1)

Recursive formula: t1 = 2, tn = 4tn-1 for n ≥ 2

Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises for Advanced GP

Questions 11 to 13 use GP conditions with sums, products and term positions. The key property is that the middle term squared equals the product of neighbouring terms.

Q11. The sum of the first three terms of a GP is 13/12 and their product is -1. Find the common ratio and the terms.

The common ratio is -4/3 or -3/4, and the terms are 3/4, -1, 4/3 or 4/3, -1, 3/4.

Let the three GP terms be:

a/r, a, ar

Product:

(a/r) × a × ar = a^3

Given:

a^3 = -1

So:

a = -1

Now the terms are:

-1/r, -1, -r

Given sum:

-1/r - 1 - r = 13/12

Multiply by 12r:

-12 - 12r - 12r^2 = 13r

Bring all terms to one side:

12r^2 + 25r + 12 = 0

Factorise:

(3r + 4)(4r + 3) = 0

So:

r = -4/3 or r = -3/4

If r = -4/3:

Terms are 3/4, -1, 4/3

If r = -3/4:

Terms are 4/3, -1, 3/4

Answer:

Common ratio = -4/3 or -3/4

Terms = 3/4, -1, 4/3 or 4/3, -1, 3/4

Q12. If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

x, y and z are in GP because y/x = z/y.

Let the GP have first term a and common ratio r.

Then:

t4 = ar^3

t10 = ar^9

t16 = ar^15

Given:

x = ar^3

y = ar^9

z = ar^15

Now:

y/x = ar^9 / ar^3

y/x = r^6

Also:

z/y = ar^15 / ar^9

z/y = r^6

So:

y/x = z/y

Therefore:

x, y, z are in GP.

Answer:

x, y and z are in GP because y/x = z/y = r^6.

Q13. The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.

The terms are 2, 6, 18 or 18, 6, 2.

Let the three GP terms be:

x, y, z

Since they are in GP:

xz = y²

Given:

x + y + z = 26

x² + y² + z² = 364

Now square the sum:

(x + y + z)² = 26²

x² + y² + z² + 2(xy + yz + zx) = 676

Substitute:

364 + 2(xy + yz + zx) = 676

2(xy + yz + zx) = 312

xy + yz + zx = 156

Since xz = y²:

xy + yz + xz = y(x + z) + y²

Also:

x + z = 26 - y

So:

y(26 - y) + y² = 156

26y = 156

y = 6

Now:

x + z = 26 - 6

x + z = 20

And:

xz = y²

xz = 36

Numbers with sum 20 and product 36 are:

2 and 18

Answer:

The terms are 2, 6, 18 or 18, 6, 2.

Class 9 Maths Sequences Solutions for Recursive Formula for Sequence

The final two questions use recursive rules built from sums of earlier terms. Both sequences simplify into familiar doubling or Fibonacci-type patterns.

Q14. Suppose P1 = 1, P2 = 2 and for n > 2, Pn = P1 + P2 + ... + Pn-1 + 1. Find P1, P2, ..., P8. Can you find a simpler recursive formula for Pn? Can you give an explicit formula?

The sequence is 1, 2, 4, 8, 16, 32, 64, 128.

Given:

P1 = 1

P2 = 2

For n > 2:

Pn = P1 + P2 + ... + Pn-1 + 1

Now:

P3 = P1 + P2 + 1

P3 = 1 + 2 + 1

P3 = 4

P4 = P1 + P2 + P3 + 1

P4 = 1 + 2 + 4 + 1

P4 = 8

Continuing:

P5 = 16

P6 = 32

P7 = 64

P8 = 128

Simpler recursive formula:

P1 = 1

Pn = 2Pn-1 for n ≥ 2

Explicit formula:

Pn = 2^(n - 1)

Answer:

P1 to P8 are 1, 2, 4, 8, 16, 32, 64, 128.

Recursive formula: P1 = 1, Pn = 2Pn-1 for n ≥ 2

Explicit formula: Pn = 2^(n - 1)

Q15. Suppose W1 = 1, W2 = 2 and for n > 2, Wn = W1 + W2 + ... + Wn-2 + 2. Find W1, W2, ..., W8. Do you recognise this sequence?

The sequence is 1, 2, 3, 5, 8, 13, 21, 34. It is the Virahanka-Fibonacci sequence pattern.

Given:

W1 = 1

W2 = 2

For n > 2:

Wn = W1 + W2 + ... + Wn-2 + 2

Now:

W3 = W1 + 2

W3 = 1 + 2

W3 = 3

W4 = W1 + W2 + 2

W4 = 1 + 2 + 2

W4 = 5

W5 = W1 + W2 + W3 + 2

W5 = 1 + 2 + 3 + 2

W5 = 8

Continuing:

W6 = 13

W7 = 21

W8 = 34

Recognised pattern:

Each term after the second is the sum of the previous two terms.

Simpler recursive formula:

W1 = 1

W2 = 2

Wn = Wn-1 + Wn-2 for n ≥ 3

Answer:

W1 to W8 are 1, 2, 3, 5, 8, 13, 21, 34.

This is the Virahanka-Fibonacci sequence pattern.

Predicting What Comes Next: Concepts Used in End of Chapter Exercises

The End of Chapter Exercises combine AP, GP, sums and recursive rules. These formulas help students solve almost every question in the exercise.

Arithmetic Progression Class 9

An arithmetic progression is a sequence with a constant difference.

Copy-friendly formula:

tn = a + (n - 1)d

Here:

a = first term

d = common difference

n = term position

Geometric Progression Class 9

A geometric progression is a sequence with a constant ratio.

Copy-friendly formula:

tn = ar^(n - 1)

Here:

a = first term

r = common ratio

n = term position

nth Term of AP Class 9

The nth term of an AP is used to find unknown terms and positions.

Copy-friendly example:

t31 = a + 30d

nth Term of GP Class 9

The nth term of a GP is used when values multiply by the same number.

Copy-friendly example:

For 2, 8, 32, ...

tn = 2 × 4^(n - 1)

Sum of Consecutive Natural Numbers

The sum of first n natural numbers is used in sum-based questions.

Copy-friendly formula:

1 + 2 + 3 + ... + n = n(n + 1)/2

For a consecutive sequence starting at m:

Sum = k/2 × [2m + k - 1]

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises

Concept	Copy-Friendly Formula	Used In
AP nth term	tn = a + (n - 1)d	Q1, Q2, Q3, Q4, Q8
GP nth term	tn = ar^(n - 1)	Q5, Q10, Q11, Q12, Q13
Consecutive sum	Sum = k/2 × [2m + k - 1]	Q6, Q9

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8

Section	NCERT Solutions
Class 9 Maths Ganita Manjari 2026	NCERT Class 9 Maths Ganita Manjari 2026
Chapter 8	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8
Exercise 8.1	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.1
Exercise 8.2	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2
Exercise 8.3	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.3
End of Chapter Exercises	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises