Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.1 Solutions

Sequence is an ordered list of numbers in which each number is called a term.
Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.1 connects sequences with nth term formulas, explicit rules, recursive rules and the Virahanka-Fibonacci type pattern.

Chapter 8, Predicting What Comes Next: Exploring Sequences and Progressions Class 9, begins with number patterns such as natural numbers, odd numbers, triangular numbers and square numbers. Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.1 Solutions focus on finding terms from an nth term rule, checking whether a number belongs to a sequence, identifying the position of a term, and using recursive rules to generate new terms. The textbook defines a sequence as an ordered list of numbers where each number is a term of the sequence.

Key Takeaways

• Sequence: An ordered list of numbers.
• nth Term: The formula used to find any term directly.
• Explicit Rule: A rule written using the term position n.
• Recursive Rule: A rule that gives a term using one or more previous terms.

Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.1 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.1 Solutions for nth Term

Exercise 8.1 begins with explicit rules. An explicit rule allows students to find the term directly by substituting the value of n.

Q1. Find the first five terms of the sequence in which the nth term is given by (i) tn = 3n - 4, (ii) tn = 2 - 5n, and (iii) tn = n² - 2n + 3 for n ≥ 1.

To find the first five terms, put n = 1, 2, 3, 4 and 5 in each formula.

Q1(i). tn = 3n - 4

For n = 1:

t1 = 3(1) - 4

t1 = 3 - 4

t1 = -1

For n = 2:

t2 = 3(2) - 4

t2 = 6 - 4

t2 = 2

For n = 3:

t3 = 3(3) - 4

t3 = 9 - 4

t3 = 5

For n = 4:

t4 = 3(4) - 4

t4 = 12 - 4

t4 = 8

For n = 5:

t5 = 3(5) - 4

t5 = 15 - 4

t5 = 11

Answer:

The first five terms are:

-1, 2, 5, 8, 11

Q1(ii). tn = 2 - 5n

For n = 1:

t1 = 2 - 5(1)

t1 = 2 - 5

t1 = -3

For n = 2:

t2 = 2 - 5(2)

t2 = 2 - 10

t2 = -8

For n = 3:

t3 = 2 - 5(3)

t3 = 2 - 15

t3 = -13

For n = 4:

t4 = 2 - 5(4)

t4 = 2 - 20

t4 = -18

For n = 5:

t5 = 2 - 5(5)

t5 = 2 - 25

t5 = -23

Answer:

The first five terms are:

-3, -8, -13, -18, -23

Q1(iii). tn = n² - 2n + 3

For n = 1:

t1 = 1² - 2(1) + 3

t1 = 1 - 2 + 3

t1 = 2

For n = 2:

t2 = 2² - 2(2) + 3

t2 = 4 - 4 + 3

t2 = 3

For n = 3:

t3 = 3² - 2(3) + 3

t3 = 9 - 6 + 3

t3 = 6

For n = 4:

t4 = 4² - 2(4) + 3

t4 = 16 - 8 + 3

t4 = 11

For n = 5:

t5 = 5² - 2(5) + 3

t5 = 25 - 10 + 3

t5 = 18

Answer:

The first five terms are:

2, 3, 6, 11, 18

Ganita Manjari Class 9 Chapter 8 Exercise 8.1: Finding Terms from Explicit Rule

The textbook explains that an explicit formula uses the position number n to calculate the value of the term. This helps find faraway terms directly without writing all earlier terms.

Q2. Find the 10th and 15th terms of the sequence tn = 5n - 3 for n ≥ 1.

The 10th term is 47, and the 15th term is 72.

Given:

tn = 5n - 3

For the 10th term, put n = 10:

t10 = 5(10) - 3

t10 = 50 - 3

t10 = 47

For the 15th term, put n = 15:

t15 = 5(15) - 3

t15 = 75 - 3

t15 = 72

Answer:

t10 = 47

t15 = 72

Class 9 Maths Chapter 8 Exercise 8.1 Solutions: Checking Terms in a Sequence

To check whether a number is a term of a sequence, set the nth term equal to that number and solve for n. If n is a natural number, the number is a term of the sequence.

Q3. Determine whether 97 and 172 are terms of the sequence tn = 5n - 3 for n ≥ 1.

97 is a term of the sequence, and 172 is also a term of the sequence.

Given:

tn = 5n - 3

Checking 97

Set:

5n - 3 = 97

5n = 97 + 3

5n = 100

n = 100/5

n = 20

Since n = 20 is a natural number, 97 is a term of the sequence.

So:

97 is the 20th term.

Checking 172

Set:

5n - 3 = 172

5n = 172 + 3

5n = 175

n = 175/5

n = 35

Since n = 35 is a natural number, 172 is a term of the sequence.

So:

172 is the 35th term.

Answer:

97 is the 20th term.

172 is the 35th term.

Class 9 Maths Ganita Manjari Chapter 8 Solutions for Term Position

A term’s position can be found by solving the nth term formula for n. The answer must be a natural number because n represents the term number.

Q4. Which term of the sequence tn = 5n - 3 for n ≥ 1 is 607?

607 is the 122nd term.

Given:

tn = 5n - 3

Set:

5n - 3 = 607

5n = 607 + 3

5n = 610

n = 610/5

n = 122

Answer:

607 is the 122nd term of the sequence.

Predicting What Comes Next Exploring Sequences and Progressions Class 9: Recursive Rule for Sequence

A recursive rule gives each term using the previous term. The textbook describes recursive rules as sequence rules that relate terms to earlier terms.

Q5. A sequence is given by the recursive rule t1 = -5, tn+1 = tn + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?

The first five terms are -5, -2, 1, 4 and 7. The number 52 is the 20th term.

Given:

t1 = -5

tn+1 = tn + 3

Find the first five terms:

t1 = -5

t2 = t1 + 3

t2 = -5 + 3

t2 = -2

t3 = t2 + 3

t3 = -2 + 3

t3 = 1

t4 = t3 + 3

t4 = 1 + 3

t4 = 4

t5 = t4 + 3

t5 = 4 + 3

t5 = 7

So the first five terms are:

-5, -2, 1, 4, 7

Now check whether 52 is a term.

This sequence starts at -5 and increases by 3 each time.

Explicit form:

tn = -5 + (n - 1) × 3

Simplify:

tn = -5 + 3n - 3

tn = 3n - 8

Set:

3n - 8 = 52

3n = 52 + 8

3n = 60

n = 20

Answer:

The first five terms are -5, -2, 1, 4 and 7.

52 is the 20th term.

Class 9 Maths Sequences Solutions: Virahanka-Fibonacci Type Recursive Rule

The final question uses a recursive rule where each new term is obtained by adding the previous three terms. This is similar to the way the Virahanka-Fibonacci sequence uses previous terms to generate new terms.

Q6. Let T1 = 1, T2 = 2, T3 = 4, and Tn = Tn-1 + Tn-2 + Tn-3 for n ≥ 4. Find T4, T5, T6, T7, and T8.

The required terms are T4 = 7, T5 = 13, T6 = 24, T7 = 44 and T8 = 81.

Given:

T1 = 1

T2 = 2

T3 = 4

Tn = Tn-1 + Tn-2 + Tn-3

For n = 4:

T4 = T3 + T2 + T1

T4 = 4 + 2 + 1

T4 = 7

For n = 5:

T5 = T4 + T3 + T2

T5 = 7 + 4 + 2

T5 = 13

For n = 6:

T6 = T5 + T4 + T3

T6 = 13 + 7 + 4

T6 = 24

For n = 7:

T7 = T6 + T5 + T4

T7 = 24 + 13 + 7

T7 = 44

For n = 8:

T8 = T7 + T6 + T5

T8 = 44 + 24 + 13

T8 = 81

Answer:

T4 = 7

T5 = 13

T6 = 24

T7 = 44

T8 = 81

Predicting What Comes Next: Concepts Used in Exercise 8.1

Exercise 8.1 introduces two ways to describe sequences: explicit rules and recursive rules. Both methods help predict terms, check term positions and continue number patterns.

Sequence Class 9

A sequence is an ordered list of numbers.

Copy-friendly example:

1, 3, 5, 7, 9, ...

Here:

t1 = 1

t2 = 3

t3 = 5

nth Term Class 9

The nth term gives the term at position n.

Copy-friendly idea:

tn = term in the nth position

Example:

If tn = 5n - 3, then:

t10 = 5(10) - 3

t10 = 47

Explicit Rule for Sequence

An explicit rule gives a term directly using n.

Copy-friendly example:

tn = 5n - 3

To find the 15th term:

t15 = 5(15) - 3

t15 = 72

Recursive Rule for Sequence

A recursive rule gives a term using a previous term or previous terms.

Copy-friendly example:

t1 = -5

tn+1 = tn + 3

This means each next term is 3 more than the previous term.

Virahanka Fibonacci Sequence Class 9

The Virahanka-Fibonacci type idea uses previous terms to make the next term.

Copy-friendly example:

V1 = 1

V2 = 2

Vn = Vn-1 + Vn-2

In Exercise 8.1 Question 6, the rule uses three previous terms:

Tn = Tn-1 + Tn-2 + Tn-3

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 8 Exercise 8.1 Solutions

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8