Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 Solutions

An arithmetic progression is a sequence in which the difference between consecutive terms remains constant.
Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 connects AP terms, recursive rules, two-digit multiples, salary increments and the sum of first n natural numbers.

Chapter 8, Predicting What Comes Next: Exploring Sequences and Progressions Class 9, develops arithmetic progressions after introducing explicit and recursive sequence rules. Exercise 8.2 focuses on finding the nth term of AP Class 9 questions, identifying a given term, writing recursive rules, solving AP word problems Class 9, and using the sum of first n natural numbers. The textbook explains that an AP can be written as a, a + d, a + 2d, a + 3d, ... and its nth term is tn = a + (n - 1)d.

Key Takeaways

• Arithmetic Progression: Consecutive terms differ by a fixed number.
• nth Term of AP: The formula is tn = a + (n - 1)d.
• Recursive Rule for AP: The rule is t1 = a, tn = tn-1 + d for n ≥ 2.
• Natural Number Sum: 1 + 2 + 3 + ... + n = n(n + 1)/2.

Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 Solutions for AP Terms

Exercise 8.2 begins with AP term-finding questions. The common difference is found by subtracting any term from the next term.

Q1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ...

The 10th term is 48, and the 26th term is 128.

Given:

First term = 3

Common difference = 8 - 3

Common difference = 5

Use:

tn = a + (n - 1)d

For the 10th term:

t10 = 3 + (10 - 1)5

t10 = 3 + 9 × 5

t10 = 3 + 45

t10 = 48

For the 26th term:

t26 = 3 + (26 - 1)5

t26 = 3 + 25 × 5

t26 = 3 + 125

t26 = 128

Answer:

t10 = 48

t26 = 128

Ganita Manjari Class 9 Chapter 8 Exercise 8.2: Finding the Position of a Term

To find which term a number is in an AP, put tn equal to that number and solve for n. Since n is the term position, the value of n must be a natural number.

Q2. Which term of the AP: 21, 18, 15, ... is -81? Also, is 0 a term of this AP? Give reasons for your answer.

-81 is the 35th term of the AP. The number 0 is the 8th term of the AP.

Given:

AP = 21, 18, 15, ...

First term = 21

Common difference = 18 - 21

Common difference = -3

Use:

tn = a + (n - 1)d

So:

tn = 21 + (n - 1)(-3)

tn = 21 - 3n + 3

tn = 24 - 3n

Checking -81

Set:

24 - 3n = -81

-3n = -81 - 24

-3n = -105

n = 35

So:

-81 is the 35th term.

Checking 0

Set:

24 - 3n = 0

3n = 24

n = 8

So:

0 is the 8th term.

Answer:

-81 is the 35th term.

0 is the 8th term of the AP.

Class 9 Maths Chapter 8 Exercise 8.2 Solutions: nth Term and Recursive Rule

An AP can be written using both an explicit rule and a recursive rule. The explicit rule gives tn directly, while the recursive rule gives each term from the previous term.

Q3. Find the nth term of the AP: 11, 8, 5, 2, ... Write the recursive rule for this AP.

The nth term is tn = 14 - 3n. The recursive rule is t1 = 11, tn = tn-1 - 3 for n ≥ 2.

Given:

AP = 11, 8, 5, 2, ...

First term = 11

Common difference = 8 - 11

Common difference = -3

Use:

tn = a + (n - 1)d

Substitute:

tn = 11 + (n - 1)(-3)

tn = 11 - 3n + 3

tn = 14 - 3n

Recursive rule:

t1 = 11

tn = tn-1 - 3 for n ≥ 2

Answer:

nth term:

tn = 14 - 3n

Recursive rule:

t1 = 11, tn = tn-1 - 3 for n ≥ 2

Class 9 Maths Ganita Manjari Chapter 8 Solutions for Linear Equations in AP

When two terms of an AP are known, the formula tn = a + (n - 1)d gives equations in a and d. Solving those equations gives the required AP.

Q4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.

The 29th term is 64.

Given:

Number of terms = 50

3rd term = 12

50th term = 106

Use:

tn = a + (n - 1)d

For the 3rd term:

t3 = a + 2d

a + 2d = 12

For the 50th term:

t50 = a + 49d

a + 49d = 106

Subtract the first equation from the second:

(a + 49d) - (a + 2d) = 106 - 12

47d = 94

d = 2

Now substitute d = 2 in:

a + 2d = 12

a + 2(2) = 12

a + 4 = 12

a = 8

Now find the 29th term:

t29 = a + (29 - 1)d

t29 = 8 + 28 × 2

t29 = 8 + 56

t29 = 64

Answer:

The 29th term is 64.

Predicting What Comes Next Exploring Sequences and Progressions Class 9: Two-Digit Multiples

Two-digit numbers divisible by 3 form an AP. The first two-digit multiple of 3 is 12, and the last two-digit multiple of 3 is 99.

Q5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

There are 30 two-digit numbers divisible by 3, and their sum is 1665.

Two-digit multiples of 3 are:

12, 15, 18, ..., 99

This is an AP.

First term:

a = 12

Common difference:

d = 3

Last term:

tn = 99

Use:

tn = a + (n - 1)d

Substitute:

99 = 12 + (n - 1)3

99 - 12 = 3(n - 1)

87 = 3(n - 1)

29 = n - 1

n = 30

So there are 30 two-digit numbers divisible by 3.

Now find the sum.

Use:

Sum = n/2 × (first term + last term)

Substitute:

Sum = 30/2 × (12 + 99)

Sum = 15 × 111

Sum = 1665

Answer:

There are 30 two-digit numbers divisible by 3.

Their sum is 1665.

Class 9 Maths Sequences Solutions: AP Word Problems Class 9

Salary increment questions are AP word problems because the salary increases by a fixed amount every year.

Q6. 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?

Harish’s income reached ₹7,00,000 after 11 years.

Given:

Starting salary = ₹5,00,000

Yearly increment = ₹20,000

Target salary = ₹7,00,000

This forms an AP:

5,00,000, 5,20,000, 5,40,000, ...

Here:

a = 5,00,000

d = 20,000

tn = 7,00,000

Use:

tn = a + (n - 1)d

Substitute:

7,00,000 = 5,00,000 + (n - 1)20,000

7,00,000 - 5,00,000 = (n - 1)20,000

2,00,000 = (n - 1)20,000

n - 1 = 10

n = 11

Answer:

Harish’s income reached ₹7,00,000 in the 11th year.

Class 9 Maths Chapter 8 Exercise 8.2 Solutions Using Sum of First n Natural Numbers

The textbook derives the formula for the sum of first n natural numbers as n(n + 1)/2. This formula is used for marble arrangements in rows.

Q7. 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?

The child uses 325 marbles in all.

The rows contain:

1, 2, 3, 4, ..., 25

Total marbles:

1 + 2 + 3 + ... + 25

Use:

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

Here:

n = 25

Substitute:

Sum = 25(25 + 1)/2

Sum = 25 × 26/2

Sum = 25 × 13

Sum = 325

Answer:

The child uses 325 marbles in all.

Arithmetic Progression Class 9: Concepts Used in Exercise 8.2

Exercise 8.2 is based on arithmetic progressions and sums. The main tools are the nth term formula, recursive rule and sum formulas.

Arithmetic Progression Class 9

An arithmetic progression is a sequence in which each term is obtained by adding a fixed number to the previous term.

Copy-friendly example:

3, 8, 13, 18, ...

Here:

Common difference = 5

nth Term of AP Class 9

The nth term of an AP gives the term at position n.

Copy-friendly formula:

tn = a + (n - 1)d

Here:

a = first term

d = common difference

n = term position

Recursive Rule for AP

A recursive rule writes each term using the previous term.

Copy-friendly formula:

t1 = a

tn = tn-1 + d for n ≥ 2

Example:

For 11, 8, 5, 2, ...

t1 = 11

tn = tn-1 - 3 for n ≥ 2

Sum of First n Natural Numbers

The sum of first n natural numbers is used when rows increase as 1, 2, 3, and so on.

Copy-friendly formula:

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

For 25 rows:

1 + 2 + 3 + ... + 25 = 25(26)/2

1 + 2 + 3 + ... + 25 = 325

Two-Digit Numbers Divisible by 3

Two-digit multiples of 3 form an AP.

Copy-friendly sequence:

12, 15, 18, ..., 99

Here:

a = 12

d = 3

last term = 99

Number of terms = 30

Sum = 1665

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.2 Solutions

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8