NCERT Solutions for Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Exercise 8.1 | 2026-27

Think and Reflect 

1. Can you describe the pattern in each of the above sequences? Can you predict the next few numbers in these sequences?

Solution:

Given sequence: 1, 4, 7, 10, 13, …

Here, the first term is 1.

Second term = 1 + 3 = 4

Third term = 4 + 3 = 7

Fourth term = 7 + 3 = 10

We can observe that each term is obtained by adding 3 to the previous term.

Therefore, the next four terms are:

13 + 3 = 16

16 + 3 = 19

19 + 3 = 22

22 + 3 = 25

Thus, the first 10 terms of the sequence are:

1, 4, 7, 10, 13, 16, 19, 22, 25, 28.

Now, let us find the sequence formed by adding all the terms up to each term of the given sequence.

First term = 1

Second term = 1 + 4 = 5

Third term = 1 + 4 + 7 = 12

Fourth term = 1 + 4 + 7 + 10 = 22

Fifth term = 1 + 4 + 7 + 10 + 13 = 35

Sixth term = 1 + 4 + 7 + 10 + 13 + 16 = 51

Seventh term = 1 + 4 + 7 + 10 + 13 + 16 + 19 = 70

Eighth term = 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 = 92

Ninth term = 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 = 117

Tenth term = 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 + 28 = 145

Hence, the first 10 terms of the new sequence are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

Think and Reflect 

1. Can you think of any other kinds of sequences? List five different types of sequences and discuss their properties with your friends.

Solution:

(i) 2, 5, 8, 11, 14, …

Add the same number, 3, to each term to get the next term.

Common difference = 3

Property: The sequence increases at a constant rate.

(ii) 3, 6, 12, 24, 48, …

Each term is multiplied by 2 to get the next term.

Common ratio = 2

Property: The sequence grows rapidly.

(iii) Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, …

Each term is obtained by adding the previous two terms.

Property: This pattern appears in many natural forms, such as the arrangement of leaves and flower petals.

(iv) 2, 3, 5, 7, 11, …

This is the sequence of prime numbers. Each number has exactly two factors: 1 and the number itself.

Property: The sequence does not follow a fixed difference or ratio.

(v) 1, 2, 6, 24, 120, …

Each term is obtained by multiplying the previous term by its position number.

Property: The sequence increases very rapidly.

Think and Reflect

1. Why is it useful to have an explicit formula for the nth term of a sequence?

Solution:

Having an explicit formula for the nth term of a sequence is useful because it allows you to find any specific term directly without having to list all the previous terms. It saves time and effort, especially when n is large. It also helps identify patterns, simplify calculations, and solve problems efficiently. Hence, an explicit formula provides a direct, quick way to find any term in the sequence.

Example:
For the sequence 2, 5, 8, 11, …, the nth term is given by:

aₙ = 2 + (n − 1) × 3

To find the 10th term:

a₁₀ = 2 + (10 − 1) × 3 = 29

So, the 10th term is 29. This shows how an explicit formula helps us find a term directly without listing all the terms before it.

Think and Reflect

1. Can you find the rule describing the nth term of the sequence of square numbers?

Solution:

The given sequence of square numbers is 1, 4, 9, 16, 25, …

Let us identify the pattern in the sequence.

Here, 1 = 1², 4 = 2², 9 = 3², 16 = 4², 25 = 5², …

Thus, each term is the square of its position number.

Let the nth term be aₙ.

∴ aₙ = n²

Hence, the rule for the nth term is aₙ = n².

Exercise Set 8.1 

1. 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. 

Solution:

(i) tn = 3n – 4,
Substitute 1, 2, 3, 4, and 5 for n in the given equation.
t1 = 3 × 1 – 4 = 3 – 4 = -1
t2 = 3 × 2 – 4 = 6 – 4 = 2
t3 = 3 × 3 – 4 = 9 – 4 = 5
t4 = 3 × 4 – 4 = 12 – 4 = 8
t5 = 3 × 5 – 4 = 15 – 4= 11
Thus, the first five terms of the given sequence are: -1, 2, 5, 8, 11

(ii) Given tn = 2 – 5n,
Substitute 1, 2, 3, 4, and 5 for n in the given equation.
t1 = 2 – 5(1) = 2 – 5 = -3
t2 = 2 – 5(2) = 2 – 10 = -8
t3 = 2 – 5(3) = 2 – 15 = -12
t4 = 2 – 5(4) = 2 – 20 = -18
t5 = 2 – 5(5) = 2 – 25 = -23
Thus, the first five terms of the given sequence: -3, -8,-13,-18, -23

(iii) Given: tn = n2 – 2n + 3
t1 = 12 – 2(1) + 3 = 2
t2 = 22 – 2(2) + 3 = 3
t3 = 32 – 2(3) + 3 = 6
t4 = 42 – 2(4) + 3 = 11
t5 = 52 – 2(5) + 3 = 18
Thus, first five terms of the given sequence are: 2, 3, 6, 11, 18

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

Solution:

Given: tn = 5n – 3

10th term means t10 i.e., n = 10

t10 = 5 × 10 – 3 = 50 – 3 = 47

15th term means t15 i.e., n = 15

t15 = 5 × 15 – 3 = 75 – 3 = 72 

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

Solution:

Given: tn = 5n – 3

Substitute 97 for tn

97 = 5n – 3

⇒ 100 = 5n

⇒ n = 20

Thus, 97 is the 20th term of the given sequence.

Now, substitute tn – 172

∴ 172 = 5n – 3

⇒ 175 = 5n

⇒ n = 35

Thus, 172 is the 35th term of the given sequence.

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

Solution:

Given: tn = 5n – 3

Substitute 607 for tn

607 = 5n — 3

⇒ 610 = 5n

Thus, 607 is 122th term of the given sequence 

5. 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? 

Solution:

Given, t1 = -5 and tn+1 = tn + 3

Put n = 1 to get 2nd term,

t1+1 = t1 + 3

⇒ t2 = -5 + 3 = -2

Put n = 2 to get 3rd term,

t2+1 = t2 + 3

⇒ t3 = -2 + 3 = 1

Put n = 3 to get 4th term,

t3+1 = t3 + 3

⇒ t4 = 1 + 3 = 4

Put n = 4 to get 5th term,

t4+1 = t4 + 3

⇒ t5 = 4 + 3 = 7

Hence, the first five terms of the sequence are -5, -2, 1, 4, 7.

Here, we get each term by adding 3 to the previous term.

Thus, the nth term of the sequence can be deduced by taking the first term and adding 3 repeatedly (n – 1) times.

i.e. tn = -5 + 3(n – 1)

= -5 + 3n – 3

= -8 + 3n

Let 52 be the nth term of this sequence.

∴ tn = 52

⇒ -8 + 3n = 52

⇒ 3n = 52 + 8

⇒ 3n = 60

⇒ n = 20

Hence, 52 is the 20th term of the sequence. 

6. 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.

Solution:

Given: T1 = 1, T2 = 2, T3 = 4

and Tn = Tn-1 +Tn-2 + Tn-3 for n > 4

Now, T4= T3 + T2 + T1

= 4 + 2 + 1

= 7

T5 = T4 + T3 + T2

= 7 + 4 + 2

= 13

T6 = T5 + T4 + T3

= 13 + 7 + 4

= 24

T7 = T6 + T5 + T4 = 24 + 13 + 7 = 44

T8 = T7 + T6 + T5 = 44 + 24 + 13 = 81

Hence, T4 = 7, T5 = 13, T6 = 24, T7 = 44, T8 = 81

How Can Exercise 8.1 from Predicting What Comes Next: Exploring Sequences and Progressions Solutions Help Class 9 Students?

Class 9 Maths Chapter 8 Exercise 8.1 Solutions help students understand the questions and the steps involved in solving them. Exercise 8.1 focuses on sequences and progressions, helping students identify patterns, understand relationships between terms and apply these ideas to different situations.

Students can first attempt each question on their own and then use the NCERT Class 9 Maths Chapter 8 Exercise 8.1 Solutions to check their answers. The step-by-step explanations can help them understand where they went wrong and learn the correct approach instead of simply memorising the answer.

Key Benefits

• Helps students understand sequences and progressions clearly

• Provides guidance for questions that seem difficult

• Helps identify and correct mistakes

• Makes revision more organised

• Improves accuracy and problem-solving skills

• Builds confidence through regular practice

How to Use These Solutions for Exam Preparation

Class 9 Maths NCERT Solutions Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Exercise 8.1 can be used effectively after completing the textbook questions. Students can compare their answers with the worked-out steps and revisit any concept they find difficult.

Students can refer to NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.1 when they need a quick explanation of a question or want to verify their final answer.

The best approach is to attempt the questions independently first, use the solutions to review the method, and then solve similar questions again without referring to the answers. This helps turn practice into stronger understanding and better exam preparation.

Access Exercise-wise NCERT Solutions for Chapter 8 Maths Class 9

CBSE Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Study Materials

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