NCERT Solutions for Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Exercise 8.2 | 2026-27

Think and Reflect 

Can you predict the number of squares in Stages 5 and 6 of the sequence? In Stages 10,11 and 12? In Stage 20? At any stage?

Solution:

From the figure, the number of squares at each stage follows the sequence 1, 5, 9, 13, …. We can see that 4 squares are added at each stage, forming the next pattern. Therefore, the number of squares in successive stages can be represented as:

1, 1 + 4, 1 + 4 + 4, 1 + 4 + 4 + 4,…

This may be written as

1 + 4 × 0, 1 + 1 × 4, 1 + 2 × 4, 1 + 3 × 4, and so on

Similarly, the number of squares in:

Stage 5: 1 + 4 × 4 = 17

Stage 6: 1 + 4 × 5 = 21

Stage 10: 1 + 4 × 9 = 37

Stage 11: 1 + 4 × 10 = 41

Stage 12: 1 + 4 × 11 = 45

Stage 20: 1 + 4 × 19 = 77

So, at any stage n, the number of sequences is given by

tn = 1 + (n – 1) × 4 = 1 + 4n – 4 = 4n – 3

Think and Reflect

1. Consider all the sequences we have discussed so far in this chapter. Which ones are arithmetic progressions and which ones are not? Can you justify your claim?

Solution:

To determine whether a sequence is an arithmetic progression, check the difference between each pair of consecutive terms. If the difference remains the same throughout, the sequence is an AP; otherwise, it is not.

Think and Reflect

1. Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?

Solution:

Let S denote the sum of the first 100 natural numbers:

S = 1 + 2 + 3 + … + 100 …(i)

Now, write the same sum in reverse order:

S = 100 + 99 + 98 + … + 1 …(ii)

Adding equations (i) and (ii), we get:

S = 1 + 2 + 3 + … + 100
S = 100 + 99 + 98 + … + 1

Adding the two equations term by term:

2S = (1 + 100) + (2 + 99) + (3 + 98) + … + (100 + 1)

Each pair of corresponding numbers adds up to 101. Since there are 100 terms, and each pair gives a sum of 101:

2S = 100 × 101 = 10100

⇒ S = 10100/2 = 5050

Therefore, the sum of the first 100 natural numbers is 5050.

Think and Reflect

1. Can you use this formula to find S20, S50 or S1000?

Solution:

We know that:

Sₙ = n(n + 1)/2

S₂₀ = 20(20 + 1)/2 = 20 × 21/2 = 210

S₅₀ = 50(50 + 1)/2 = 50 × 51/2 = 1275

And S₁₀₀₀ = 1000(1000 + 1)/2 = 1000 × 1001/2 = 500500

Think and Reflect

Let us revisit the sequence tₙ of triangular numbers 1, 3, 6, 10, 15, … shown in the Figure. Note that the nth term of this sequence is the sum of the first n natural numbers. Thus,

tₙ = n(n + 1)/2

Can you use this to find the 10th, 17th and 80th triangular numbers?

Solution:

Given: The nth term of the triangular number sequence is:

tₙ = n(n + 1)/2

10th triangular number:
Substitute n = 10 into the given formula.

t₁₀ = 10(10 + 1)/2
= 10 × 11/2
= 55

17th triangular number:
Substitute n = 17 into the given formula.

t₁₇ = 17(17 + 1)/2
= 17 × 18/2
= 153

80th triangular number:
Substitute n = 80 into the given formula.

t₈₀ = 80(80 + 1)/2
= 80 × 81/2
= 3240

Therefore, the 10th, 17th, and 80th triangular numbers are 55, 153, and 3240, respectively.

Exercise Set 8.2

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

Solution:

The given arithmetic progression (AP) is: 3, 8, 13, 18,…
Here, the first term (a) = 3
and the common difference (d) = 8 – 3 = 5.
The formula for the 77th term of an AP is given by: tn = a + (n – 1).d

For the 10th term:
Substitute n = 10. a = 3, and d = 5 into the formula:
t10= 3 + (10 – 1) × 5
= 3 + 9 × 5
= 3 + 45
= 48
Thus, the 10th term is 48.

For the 26th term:
Substitute n = 26, a = 3, and d = 5 into the formula:
t26 = 3 + (26 – 1) × 5
= 3 + 25 × 5
= 3 + 125
= 128
Thus, the 26th term is 128.

2. 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, the AP is 21, 18, 15,

Here, the first term, a = t1 = 21

and common difference, d = t2 – t1

= 18 – 21

= -3

General term of an AP is given by tn = a + (n – 1)d

Let -81 be the nth term of the AP.

∴ -81 = a + (n – 1)d

⇒ 81 = 21 + (n – 1)(-3) [∵ a = 21 and d = -3]

⇒ -81 – 21 = (n – 1)(-3)

⇒ -102 = (n – 1)(-3)

⇒ n – 1 = 34

⇒ n = 34 + 1

⇒ n = 35

Therefore, -81 is the 35th term of the AP.

Let 0 be the mth term of the AP.

0 = a + (m – 1)d

⇒ 0 = 21 + (m – 1)(-3) [∵ a = 21 and d = -3]

⇒ -21 = (m – 1)(-3)

⇒ m – 1 = 7

⇒ m = 7 + 1 = 8

Yes, 0 is the 8th term of the AP. 

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

Solution:

The given AP is: 11, 8, 5, 2,…

Here, a = 11 and d = 8 – 11 = -3.

The formula for the nth term of an AP is given by: t_n = a + (n – \).d

Substitute a = 11 and d = – 3 into the formula:

tn = 11 + (n – 1) × (-3)

= 11 – 3n + 3

= 14 – 3n

Thus, the 77th term of the AP is:

tn = 14 – 3n

⇒ t1 = 11

t2 = 14 – 6 = 8 = 11 – 3

t3 = 14 – 9 = 5 = 8 – 3

∴ The recursive rule for this AP is: t1 = 11

tn = tn-1 + (-3) = tn-1 – 3 for n ≥ 2.  

4. An AP consists of 50 terms in which the 3rd term is 12, and the last term is 106. Find the 29th term. (Hint: If ‘a’ is the first term and ‘d’ is 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:

Given, no. of terms, n = 50

t3 = 12

then, a + (3 – 1)d = 12

a + 2d = 12 ……(1)

Again t50 = 106

⇒ a + (50 – 1)d = 106

⇒ a + 49d = 106

⇒ a = 106 – 49d …..(2)

Put the value of ‘a’ obtained from (2) into (1)

106 – 49d + 2d = 12

⇒ 106 – 47d = 12

⇒ 106 – 12 = 47d

⇒ 94 = 47d

⇒ 2 = d

Put d = 2 in equation (1)

a + 2(2) = 12

⇒ a = 12 – 4

⇒ a = 8

Hence, t29 = a + (29 – 1)d

= 8 + 28 × 2

= 8 + 56

t29 = 64 

5. 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 form an arithmetic progression, starting with 12. And the sequence is: 12, 15, 18,…, 99.

The first term (a) = 12,

the last term (l) = 99, and

common difference (d) = 3.

We use the formula for the nth term of an AP:

tn = a + (n – 1 )d

Substitute tn = 99, a = 12, and d = 3 into the formula:

99 = 12 + (n – 1) × 3

⇒ 99 – 12 = (n – 1) × 3

⇒ 87 = (n – 1) × 3

⇒n – 1 = 87/3

= 29

⇒ n = 29 + 1 = 30

Thus, there are 30 terms.

Since there are 30 terms in the sequence, we can form 15 pairs.

The sum of each pair is (12 + 99) + (15 + 96) + (18 + 93) + …

Each pair sums to 111, and there are 15 such pairs.

Thus, the total sum is: 15 × 111 = 1665

Therefore, the sum of all the 2-digit numbers divisible by 3 is 1665. 

6. 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:

Given, Harish’s annual salary = ₹ 500000

and salary increment each year = ₹ 20000.

Since each year’s salary increases by a fixed amount,

So, this situation forms an AP, with a1 = 500000 and d = 20000.

Let his salary reach ₹ 700000 after n years.

∴ tn = 700000

⇒ a + (n – 1)d = 700000

⇒ 500000 + (n – 1) (20000) = 700000

⇒ (n – 1) (20000) = 700000 – 500000

⇒ (n – 1) (20000) = 200000

⇒ n – 1 = 10

⇒ n = 11

Hence, the salary becomes ₹ 700000 in the 11th year.

7. 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 each row forms an arithmetic progression.

The first term (a) = 1, the common difference (d) = 1, number of terms (n) = 25

So the arithmetic progression is: 1 + 2 + 3 + …+ 25

This is the sum of the first 25 natural numbers.

The sum of the first n natural numbers is given by: n(n+1)/2

Substitute n = 25,

S25 = 25(25+1)2=25×262 = 325

Thus, the child uses 325 marbles in all.

How Can Class 9 Maths Chapter 8 Exercise 8.2 Solutions Help Students?

The NCERT Class 9 Maths Chapter 8 Exercise 8.2 Solutions help students understand arithmetic progressions and related patterns through clear, step-by-step explanations. The exercise includes questions on finding terms of an AP, checking whether a number belongs to an AP, finding unknown terms and applying arithmetic progression concepts to different situations.

Students can first attempt each question independently and then refer to the solutions to check their method and final answer. The Class 9 Maths NCERT Solutions Chapter 8, Predicting What Comes Next: Exploring Sequences and Progressions, Exercise 8.2 can help them identify mistakes, understand the reasoning behind each step, and apply the same approach to similar questions.

Key Benefits

• Helps students understand arithmetic progressions and sequences

• Provides step-by-step guidance for difficult questions

• Makes it easier to check answers and identify mistakes

• Helps students practise different types of AP questions

• Supports revision before school and CBSE examinations

• Builds confidence in applying formulas to numerical problems

You can also use the CBSE Class 9 Maths Chapter 8 Exercise 8.2 Solutions during revision to review key question types and strengthen problem-solving skills.

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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