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Class 9 Ganita ManjariClass 9 Maths Solutions | 2026–27
Chapter 8: Predicting What Comes Next: Exploring Sequences and ProgressionsPage 185–186All 7 Questions on This Page

Class 9 Maths Chapter 8 Exercise Set 8.2 Solutions

Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 8 Exercise Set 8.2. Covers finding nth terms of an AP, identifying term positions, recursive rules, solving simultaneous linear equations for AP parameters, and AP sum applications.

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Key Concepts for Exercise Set 8.2

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1

Arithmetic Progression (AP)

A sequence of numbers where the difference between any consecutive terms remains constant throughout.

2

First Term (a)

The very first term of the arithmetic progression, denoted by a or t₁.

t₁ = a
3

Common Difference (d)

The constant difference added to each term to obtain the next term (d = tₙ − tₙ₋₁). Can be positive, negative, or zero.

d = tₙ − tₙ₋₁
4

General nth Term (tₙ)

Used to determine any term position in an AP without writing out all preceding terms:

tₙ = a + (n − 1)d
5

Finding Term Position (n)

To find which term has a given value, set tₙ equal to that value and solve the linear equation for n. The value of n must always be a positive integer (n ∈ {1, 2, 3, …}).

6

Recursive Rule for an AP

Expresses each term based on the value of the term immediately before it:

t₁ = a, tₙ = tₙ₋₁ + d, for n ≥ 2
7

Finding a and d from Given Terms

When two distinct terms are known (e.g. t₃ and t₅₀), form a pair of simultaneous linear equations using tₙ = a + (n − 1)d and solve for a and d.

t₃: a + 2d = 12, t₅₀: a + 49d = 106
8

Sum of First n Terms (Sₙ)

When the first term a and the last term l are known, the sum of all n terms is given by:

Sₙ = n/2 (a + l)

Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….

Solution
Given:
a = 3
d = 8 − 3 = 5
We know,
tₙ = a + (n − 1)d
For the 10th term:

t₁₀ = 3 + (10 − 1) × 5

= 3 + 45
= 48
For the 26th term:

t₂₆ = 3 + (26 − 1) × 5

= 3 + 125
= 128
Therefore:
t₁₀ = 48
t₂₆ = 128
Answer:t₁₀ = 48 and t₂₆ = 128

Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer.

Solution
Given:
a = 21
d = 18 − 21 = −3
We know,
tₙ = a + (n − 1)d
For tₙ = −81:
−81 = 21 + (n − 1)(−3)
−81 = 21 − 3(n − 1)
−102 = −3(n − 1)
n − 1 = 34
n = 35

So, −81 is the 35th term.

To check whether 0 is a term:
0 = 21 − 3(n − 1)
−21 = −3(n − 1)
n − 1 = 7
n = 8

Therefore, 0 is the 8th term.

Final result:

−81 is the 35th term, and 0 is the 8th term.

Answer:−81 is the 35th term, and 0 is the 8th term.

Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.

Solution
Given:
a = 11
d = 8 − 11 = −3
Using:
tₙ = a + (n − 1)d
we get,
tₙ = 11 + (n − 1)(−3)
= 11 − 3n + 3
= 14 − 3n
Therefore:
tₙ = 14 − 3n
Recursive rule:

The first term is:

t₁ = 11

Each term is obtained by subtracting 3 from the previous term:

tₙ = tₙ₋₁ − 3, for n ≥ 2

Answer:nth term is tₙ = 14 − 3n; Recursive rule: t₁ = 11, tₙ = tₙ₋₁ − 3 for n ≥ 2.

An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.

💡Hint from textbook:

If ‘a’ is the first term and ‘d’ the common difference, then we arrive at the equations a + 2d = 12 and a + 49d = 106. Solve this pair of linear equations for ‘a’ and ‘d’.

Solution

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

We know,
tₙ = a + (n − 1)d
Given:
t₃ = 12
a + 2d = 12--- (1)

and the last term (50th term):

t₅₀ = 106
a + 49d = 106--- (2)
Subtracting (1) from (2):

(a + 49d) − (a + 2d) = 106 − 12

47d = 94
d = 2
Substituting d = 2 in (1):
a + 2(2) = 12
a + 4 = 12
a = 8
For the 29th term:
t₂₉ = a + (29 − 1)d
t₂₉ = 8 + 28 × 2
= 8 + 56
= 64
Therefore:
t₂₉ = 64
Answer:t₂₉ = 64

How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

Solution

The 2-digit numbers divisible by 3 are:

12, 15, 18, …, 99

This is an AP with:

a = 12
d = 3
l = 99
Using:
l = a + (n − 1)d
we get,
99 = 12 + (n − 1)3
87 = 3(n − 1)
29 = n − 1
n = 30
So, there are 30 two-digit numbers divisible by 3.

Their sum is:

Sₙ = n/2 (a + l)
we get,
S₃₀ = 302 (12 + 99)
= 15 × 111
= 1665
Therefore:
Number of two-digit numbers = 30
Sum S₃₀ = 1665
Answer:Number of two-digit numbers = 30; Sum S₃₀ = 1665

Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

Solution

The yearly salaries form an AP:

₹5,00,000, ₹5,20,000, ₹5,40,000, …

Here:

a = ₹5,00,000
d = ₹20,000

We want the salary to reach ₹7,00,000.

Using:
tₙ = a + (n − 1)d
we get,
7,00,000 = 5,00,000 + (n − 1)(20,000)
2,00,000 = (n − 1)(20,000)
10 = n − 1
n = 11

Therefore, ₹7,00,000 is the 11th year's salary.

Since the starting salary is the first year's salary:

Number of years = 11 − 1 = 10

Therefore, his income reached that amount after 10 years.
Answer:After 10 years (as ₹7,00,000 is the salary for the 11th year, t₁₁).

A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?

Solution

The number of marbles in the rows is:

1, 2, 3, …, 25

This is an AP with:

a = 1
l = 25
n = 25
Total marbles:
Sₙ = n/2 (a + l)
we get,
S₂₅ = 252 (1 + 25)
= 252 × 26
= 25 × 13
= 325
Therefore:

325 marbles

Answer:Total marbles used S₂₅ = 325

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