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Class 9 Ganita ManjariClass 9 Maths Solutions | 2026–27
Chapter 8: Predicting What Comes Next: Exploring Sequences and ProgressionsPage 194–195All 15 Questions on This Page

Class 9 Maths Chapter 8 End-of-Chapter Exercises Solutions

Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 8 End-of-Chapter Exercises (Page 194–195). Comprehensive chapter revision covering Arithmetic Progressions (AP), Geometric Progressions (GP), evaluating specific terms, finding term positions, expressing numbers as sums of consecutive integers, recursive and explicit sequence rules, and the Virahānka-Fibonacci sequence.

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Key Concepts for End-of-Chapter Exercises

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1

Arithmetic Progression (AP)

A sequence where each term after the first is obtained by adding a fixed number called the common difference (d) to the preceding term.

a, a + d, a + 2d, a + 3d, …, a + (n − 1)d
2

Geometric Progression (GP)

A sequence of non-zero terms where each term after the first is obtained by multiplying the previous term by a fixed non-zero number called the common ratio (r).

a, ar, ar², ar³, …, arⁿ⁻¹
3

General nth Term of an AP

Formula for finding the term at position n in an AP with first term a and common difference d.

tₙ = a + (n − 1)d
4

General nth Term of a GP

Formula for finding the term at position n in a GP with first term a and common ratio r.

tₙ = arⁿ⁻¹
5

Sum of First n Natural Numbers

Formula for evaluating the sum of the first n consecutive natural numbers 1 + 2 + 3 + … + n.

Sₙ = n(n + 1) / 2
6

Sum of Consecutive Terms in an AP

The sum of n terms of an arithmetic progression with first term a and common difference d.

S = n/2 [2a + (n − 1)d]
7

Explicit vs Recursive Sequence Formulas

An explicit formula computes tₙ directly from n. A recursive rule expresses tₙ in terms of preceding terms along with initial base values.

Explicit: tₙ = f(n) | Recursive: t₁ = a, tₙ = g(tₙ₋₁)
8

Virahānka-Fibonacci Sequence

A sequence where each term after the first two is the sum of the two immediately preceding terms: 1, 2, 3, 5, 8, 13, 21, 34, …

W₁ = 1, W₂ = 2, Wₙ = Wₙ₋₁ + Wₙ₋₂ (n ≥ 3)
Question 1
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Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Solution

Let the first term be a and the common difference be d.

We know,
tₙ = a + (n − 1)d
Given,
t₁₁ = 38
a + 10d = 38
Also,
t₁₆ = 73
a + 15d = 73
Subtracting,

(a + 15d) − (a + 10d) = 73 − 38

5d = 35
d = 7
Now,
a + 10(7) = 38
a + 70 = 38
a = −32
For the 31st term:
t₃₁ = a + 30d
= −32 + 30(7)
= −32 + 210
= 178
Therefore,
t₃₁ = 178
Answer:t₃₁ = 178
Question 2
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Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

Solution

Let the first term be a and the common difference be d.

We know,
tₙ = a + (n − 1)d
Given,
t₃ = 16
a + 2d = 16
Also,
t₇ − t₅ = 12

(a + 6d) − (a + 4d) = 12

2d = 12
d = 6
Now,
a + 2(6) = 16
a + 12 = 16
a = 4
Therefore, the AP is:
4, 10, 16, 22, …
Answer:4, 10, 16, 22, …
Question 3
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How many three-digit numbers are divisible by 7?

💡Hint from textbook:

All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.

Solution

The three-digit numbers divisible by 7 form an AP.

The smallest three-digit number divisible by 7 is:
105
The largest three-digit number divisible by 7 is:
994
So,

a = 105, tₙ = 994, d = 7

Using,
tₙ = a + (n − 1)d
we get,
994 = 105 + (n − 1)7
889 = 7(n − 1)
127 = n − 1
n = 128
Therefore,
128

three-digit numbers are divisible by 7.

Answer:128 three-digit numbers are divisible by 7.
Question 4
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How many multiples of 4 lie between 10 and 250?

💡Hint from textbook:

All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250.

Solution

The multiples of 4 between 10 and 250 are:

12, 16, 20, …, 248

The smallest multiple of 4 between 10 and 250 is:
12
The largest multiple of 4 between 10 and 250 is:
248
So,

a = 12, tₙ = 248, d = 4

Using,
tₙ = a + (n − 1)d
we get,
248 = 12 + (n − 1)4
236 = 4(n − 1)
59 = n − 1
n = 60
Therefore,
60

multiples of 4 lie between 10 and 250.

Answer:60 multiples of 4
Question 5
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Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.

Solution

Let a be the first term and r be the common ratio of the GP.

We know,
tₙ = arⁿ⁻¹
Given,
t₁ + t₂ = −4
a + ar = −4
a(1 + r) = −4
Also,
t₅ = 4t₃
ar⁴ = 4ar²
r² = 4
Therefore,
r = 2 or r = −2
Case 1: r = 2
a(1 + 2) = −4
3a = −4
a = −43
Therefore, one GP is:
−43, −83, −163, …
Case 2: r = −2
a(1 − 2) = −4
−a = −4
a = 4
Therefore, another GP is:
4, −8, 16, …
Hence, the required GPs are:
−43, −83, −163, …

or

4, −8, 16, …
Answer:−43, −83, −163, … or 4, −8, 16, …
Question 6
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Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Solution

Let the number of consecutive natural numbers be n, starting from a.

For n consecutive natural numbers,

100 = n/2 [2a + (n − 1)]

So,
200 = n[2a + (n − 1)]

We check the possible values of n for which a is a natural number.

For n = 1,
100 = 12 [2a]
100 = a
Therefore,
100

is one way.

For n = 5,
100 = 52 [2a + 4]
40 = 2a + 4
2a = 36
a = 18
Therefore,

100 = 18 + 19 + 20 + 21 + 22

For n = 8,
100 = 82 [2a + 7]
25 = 2a + 7
2a = 18
a = 9
Therefore,

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

Hence, the possible ways are:
100

or

18 + 19 + 20 + 21 + 22

or

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

Answer:100 or 18 + 19 + 20 + 21 + 22 or 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16
Question 7
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The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?

Solution
Given:

Original number of bacteria at 0 hours = 30

The number of bacteria doubles every hour.

At the end of the 1st hour:
30 × 2 = 60
At the end of the 2nd hour:
60 × 2 = 120
Therefore, the sequence of the number of bacteria at the end of each hour is:
60, 120, 240, 480, …

For this GP,

a = 60
r = 2
For the end of the 2nd hour:
t₂ = 60 × 2²⁻¹
= 60 × 2
= 120
For the end of the 4th hour:
t₄ = 60 × 2⁴⁻¹
= 60 × 2³
= 60 × 8
= 480
For the end of the nth hour:
tₙ = 60 × 2ⁿ⁻¹
= (30 × 2) × 2ⁿ⁻¹
= 30 × 2ⁿ
Therefore,
End of 2nd hour = 120
End of 4th hour = 480

End of nth hour = 30 × 2ⁿ

Answer:2nd hour = 120 | 4th hour = 480 | nth hour = 30 × 2ⁿ
Question 8
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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.

Solution

Let the first term be a and the common difference be d.

We know,
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
Subtracting,

(a + 7d) − (a + 5d) = 22 − 12

2d = 10
d = 5
Now,
a + 5(5) = 12
a + 25 = 12
a = −13
Therefore, the first three terms are:
−13, −8, −3
Answer:−13, −8, −3
Question 9
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Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Solution

The first n natural numbers are:

1, 2, 3, …, n

Their sum is:

Sₙ = nn + 12
We need,
nn + 12 > 1000
Therefore,
n(n + 1) > 2000
Now, check n = 44:
44(44 + 1) = 44 × 45
= 1980
Since,
1980 < 2000

n = 44 does not satisfy the condition.

Now, check n = 45:
45(45 + 1) = 45 × 46
= 2070
Since,
2070 > 2000

the required value is:

45
Therefore, the smallest value of n is 45.
Answer:n = 45
Question 10
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Which term of the GP: 2, 8, 32, ... is 131072? Write the explicit formula as well as the recursive formula for the nth term.

Solution

The GP is:

2, 8, 32, …

Here,

a = 2
r = 82
= 4

For a GP,

tₙ = arⁿ⁻¹
Which term is 131072?
2 × 4ⁿ⁻¹ = 131072
4ⁿ⁻¹ = 1310722
4ⁿ⁻¹ = 65536
4ⁿ⁻¹ = 4⁸(since, 65536 = 4⁸)
Therefore,
n − 1 = 8
n = 9

So, 131072 is the 9th term.

Explicit formula:
tₙ = 2 × 4ⁿ⁻¹
Recursive formula:
t₁ = 2
tₙ = 4tₙ₋₁, for n ≥ 2
Therefore,

131072 is the 9th term.

Answer:131072 is the 9th term | Explicit: tₙ = 2 × 4ⁿ⁻¹ | Recursive: t₁ = 2, tₙ = 4tₙ₋₁ (n ≥ 2)
Question 11
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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.

Solution

Let the first three terms be:

a/r, a, ar

where a is the middle term and r is the common ratio.

Their product is −1:

(a/r) × a × (ar) = −1
a³ = −1
a = −1
Also,
a/r + a + ar = 1312
Substituting a = −1:
−1/r − 1 − r = 1312
12(r² + r + 1) = −13r
12r² + 25r + 12 = 0
(4r + 3)(3r + 4) = 0
Therefore,
r = −34

or

r = −43
Case 1: r = −3/4
First term:
a/r = (−1)/(−34)
= 43
Second term:
a = −1
Third term:
ar = (−1)(−34)
= 34
So the terms are:
43, −1, 34
Case 2: r = −4/3
First term:
a/r = (−1)/(−43)
= 34
Second term:
a = −1
Third term:
ar = (−1)(−43)
= 43
So the terms are:
34, −1, 43
Therefore,
r = −34 or −43

and the terms are:

43, −1, 34

or

34, −1, 43
Answer:r = −34 or −43; Terms: 43, −1, 34 or 34, −1, 43
Question 12
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If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y and z are in GP.

Solution

Let a be the first term and r be the common ratio.

We know,
tₙ = arⁿ⁻¹
Given,
x = t₄ = ar³
y = t₁₀ = ar⁹
z = t₁₆ = ar¹⁵
Now,
y/x = ar⁹ / ar³
= r⁶
Also,
z/y = ar¹⁵ / ar⁹
= r⁶
Therefore,
y/x = z/y
Hence, the ratio between consecutive terms is the same.
Therefore,

x, y and z are in GP.

Answer:Since y/x = z/y = r⁶, the ratio between consecutive terms is equal. Hence, x, y and z are in GP.
Question 13
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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.

Solution

Let the first three terms be:

a, ar, ar²

Given,
a + ar + ar² = 26

a(1 + r + r²) = 26 (1)

Also,
a² + (ar)² + (ar²)² = 364

a²(1 + r² + r⁴) = 364 (2)

Squaring equation (1),

a²(1 + r + r²)² = 676 (3)

Dividing equation (3) by equation (2),

[(1 + r + r²)²] / (1 + r² + r⁴)

= 676364
= 137
Using the identity,
1 + r² + r⁴

= (1 + r + r²)(1 − r + r²)

we get,

(1 + r + r²)/(1 − r + r²)

= 137
Cross-multiplying,

7(1 + r + r²) = 13(1 − r + r²)

7 + 7r + 7r² = 13 − 13r + 13r²

6r² − 20r + 6 = 0

Dividing by 2,

3r² − 10r + 3 = 0
(3r − 1)(r − 3) = 0
Therefore,
r = 3

or

r = 13
Case 1: r = 3

From equation (1),

a(1 + 3 + 9) = 26
13a = 26
a = 2
Therefore, the terms are:
2, 6, 18
Case 2: r = 1/3

From equation (1),

a(1 + 13 + 19) = 26
a(139) = 26
a = 18
Therefore, the terms are:
18, 6, 2
Hence, the terms of the GP are:
2, 6, 18

or

18, 6, 2
Answer:2, 6, 18 or 18, 6, 2
Question 14
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Suppose P₁ = 1, P₂ = 2 and for n > 2, Pₙ = P₁ + P₂ + ··· + Pₙ₋₁ + 1 Find the values of P₁, P₂, …, P₈. Can you find a simpler recursive formula for Pₙ? Can you give an explicit formula?

Solution
Given,
P₁ = 1, P₂ = 2

For P₃,

P₃ = P₁ + P₂ + 1
= 1 + 2 + 1
= 4

For P₄,

P₄ = P₁ + P₂ + P₃ + 1

= 1 + 2 + 4 + 1

= 8

Similarly,

P₅ = 1 + 2 + 4 + 8 + 1 = 16

P₆ = 1 + 2 + 4 + 8 + 16 + 1 = 32

P₇ = 1 + 2 + 4 + 8 + 16 + 32 + 1 = 64

P₈ = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1 = 128

Therefore,

P₁, P₂, …, P₈

= 1, 2, 4, 8, 16, 32, 64, 128

Simpler recursive formula
Notice,
P₂ = 2P₁
P₃ = 2P₂
P₄ = 2P₃

and so on.

Therefore, the simpler recursive formula is:
P₁ = 1,
Pₙ = 2Pₙ₋₁, n ≥ 2
Explicit formula

Since each term is multiplied by 2,

Pₙ = 2ⁿ⁻¹
Therefore,
Pₙ = 2ⁿ⁻¹
Answer:P₁, P₂, …, P₈: 1, 2, 4, 8, 16, 32, 64, 128 | Recursive: P₁ = 1, Pₙ = 2Pₙ₋₁ (n ≥ 2) | Explicit: Pₙ = 2ⁿ⁻¹
Question 15
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Suppose W₁ = 1, W₂ = 2 and for n > 2, Wₙ = W₁ + W₂ + … + Wₙ₋₂ + 2. Find the values of W₁, W₂, …, W₈. Do you recognise this sequence?

Solution
Given:
W₁ = 1
W₂ = 2
W₃ = W₁ + 2
= 1 + 2
= 3
W₄ = W₁ + W₂ + 2
= 1 + 2 + 2
= 5

W₅ = W₁ + W₂ + W₃ + 2

= 1 + 2 + 3 + 2

= 8

W₆ = 1 + 2 + 3 + 5 + 2

= 13

W₇ = 1 + 2 + 3 + 5 + 8 + 2

= 21

W₈ = 1 + 2 + 3 + 5 + 8 + 13 + 2

= 34
Therefore,

W₁, W₂, …, W₈:

1, 2, 3, 5, 8, 13, 21, 34

Notice:
W₃ = W₂ + W₁
W₄ = W₃ + W₂
W₅ = W₄ + W₃

and so on.

Therefore,

Wₙ = Wₙ₋₁ + Wₙ₋₂, for n ≥ 3

This is the Virahānka-Fibonacci sequence.

Answer:W₁, W₂, …, W₈: 1, 2, 3, 5, 8, 13, 21, 34; Virahānka-Fibonacci sequence

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