Important Questions Class 9 Maths Chapter 8 Sequences and Progressions 2026-2027

Important Questions Class 9 Maths Chapter 8 Sequences and Progressions

These Important Questions Class 9 Maths Chapter 8 help you practise number patterns. The questions cover finding the next terms of a sequence, recursive and explicit formulas, the sum of the first n natural numbers, arithmetic progressions (AP) and geometric progressions (GP), and a real-life question on bacteria that double every hour.

An AP is a list of numbers where we add the same number each time. A GP is a list of numbers where we multiply by the same number each time. Every answer below is solved step by step, so you can follow each line and check your own working.

Q1. The sum of the first n natural numbers is given by the formula S_n = n(n + 1)/2. Using this formula, find the sums of the first 20, 50 and 1000 natural numbers.

Answer: Given:

S_n = n(n + 1)/2

(a) Sum of the first 20 natural numbers:

S₂₀ = (20 × 21)/2

= 210

(b) Sum of the first 50 natural numbers:

S₅₀ = (50 × 51)/2

= 1275

(c) Sum of the first 1000 natural numbers:

S₁₀₀₀ = (1000 × 1001)/2

= 500500

Q2. A sequence is given by the recursive rule t₁ = −5, t_n+1 = t_n + 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?

Answer: Given: t₁ = −5 and t_n+1 = t_n + 3

First five terms:

t₁ = −5

t₂ = t₁ + 3 = −5 + 3 = −2 [Using t_n+1 = t_n + 3]

t₃ = t₂ + 3 = −2 + 3 = 1

t₄ = t₃ + 3 = 1 + 3 = 4

t₅ = t₄ + 3 = 4 + 3 = 7

Therefore, the first five terms: −5, −2, 1, 4, 7

Now checking whether 52 is a term:

Each term is 3 more than the one before, so t_n = −5 + (n − 1) × 3 = 3n − 8

Let t_n = 52

⇒ 3n − 8 = 52

⇒ 3n = 60

⇒ n = 20

Since n = 20 is a natural number, 52 is a term of the sequence. Therefore, 52 is the 20^th term.

Q3. Observe the following sequences. Identify the pattern in each sequence and write the next three terms.

1. 1, 2, 3, 4, 5, 6, …

2. 1, 3, 5, 7, 9, 11, …

3. 1, 3, 6, 10, 15, 21, …

4. 1, 4, 9, 16, 25, 36, …

Answer: Each term increases by 1.

Next three terms: 7, 8, 9

Each term increases by 2.

Next three terms: 13, 15, 17

Consecutive natural numbers are added successively: +2, +3, +4, +5, +6, …

Next three terms: 28, 36, 45

The terms are squares of consecutive natural numbers: 1², 2², 3², 4², …

Next three terms: 49, 64, 81

Q4. Suppose W₁ = 1, W₂ = 2 and for n > 2, W_n = W₁ + W₂ + … + W_n−2 + 2. Find the values of W₁, W₂, …, W₈. Do you recognise this sequence?

Answer: Given:

W₁ = 1

W₂ = 2

For n > 2: W_n = W₁ + W₂ + … + W_n−2 + 2

Now: W₃ = W₁ + 2

= 1 + 2

= 3

W₄ = W₁ + W₂ + 2

= 1 + 2 + 2

= 5

W₅ = W₁ + W₂ + W₃ + 2

= 1 + 2 + 3 + 2

= 8

W₆ = W₁ + W₂ + W₃ + W₄ + 2

= 1 + 2 + 3 + 5 + 2

= 13

W₇ = 1 + 2 + 3 + 5 + 8 + 2

= 21

W₈ = 1 + 2 + 3 + 5 + 8 + 13 + 2

= 34

Therefore: W₁, W₂, …, W₈ are 1, 2, 3, 5, 8, 13, 21, 34

Yes, this is the Virahanka–Fibonacci sequence.

Q5. Suppose P₁ = 1, P₂ = 2 and for n > 2, P_n = P₁ + P₂ + … + P_n−1 + 1. Find the values of P₁, P₂, …, P₈. Can you find a simpler recursive formula for P_n? Can you give an explicit formula?

Answer: Given:

P₁ = 1

P₂ = 2

For n > 2:

P_n = P₁ + P₂ + … + P_n−1 + 1

Now:

P₃ = P₁ + P₂ + 1

= 1 + 2 + 1

= 4

P₄ = P₁ + P₂ + P₃ + 1

= 1 + 2 + 4 + 1

= 8

P₅ = 1 + 2 + 4 + 8 + 1

= 16

P₆ = 1 + 2 + 4 + 8 + 16 + 1

= 32

P₇ = 64

P₈ = 128

So: P₁, P₂, …, P₈ are 1, 2, 4, 8, 16, 32, 64, 128

Simpler recursive formula:

P₁ = 1

P_n = 2P_n−1, for n ≥ 2

Explicit formula: P_n = 2ⁿ⁻¹.

Q6. The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.

Answer: Let the three terms be a, ar, ar²

According to first condition:

a + ar + ar² = 26

⇒ a(1 + r + r²) = 26 …(1)

According to second condition:

(a)² + (ar)² + (ar²)² = 364

⇒ a²(1 + r² + r⁴) = 364 …(2)

Now, dividing equation (2) by square of equation (1), we have

a²(1 + r² + r⁴) / [a(1 + r + r²)]² = 364/26²

⇒ (1 + r² + r⁴)/(1 + r + r²)² = 7/13

⇒ 13(1 + r² + r⁴) = 7(1 + r + r²)²

⇒ 13(1 + r² + r⁴) − 7(r² + r + 1)² = 0

⇒ 13(r² + r + 1)(r² − r + 1) − 7(r² + r + 1)² = 0 [Using 1 + r² + r⁴ = (r² + r + 1)(r² − r + 1)]

⇒ [13(r² − r + 1) − 7(r² + r + 1)] = 0 [Dividing by (r² + r + 1)]

⇒ [13r² − 13r + 13 − 7r² − 7r − 7] = 0

⇒ (6r² − 20r + 6) = 0

⇒ 3r² − 10r + 3 = 0

⇒ (3r − 1)(r − 3) = 0

∴ r = 1/3 or r = 3

If r = 1/3, from (1), we have

a[1 + 1/3 + (1/3)²] = 26

⇒ a[1 + 1/3 + 1/9] = 26

⇒ a[(9 + 3 + 1)/9] = 26

⇒ a[13/9] = 26

⇒ a = 18

Hence, the three terms a, ar, ar² are 18, 6, 2.

If r = 3, from (1), we have

a[1 + 3 + (3)²] = 26

⇒ a[1 + 3 + 9] = 26

⇒ a[13] = 26

⇒ a = 2

Hence, the three terms a, ar, ar² are 2, 6, 18.

So, the terms of the GP are 2, 6, 18.

Q7. If the 4^th, 10^th and 16^th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

Answer: Let the first term of the GP be a and common ratio be r.

Then:

4^th term = ar³ = x

10^th term = ar⁹ = y

16^th term = ar¹⁵ = z

Now: y/x = ar⁹/ar³ = r⁶

Also: z/y = ar¹⁵/ar⁹ = r⁶

Therefore: y/x = z/y

So, x, y, z are in GP.

Hence proved.

Q8. Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the n^th term.

Answer: Given GP: 2, 8, 32, …

First term, a = 2

Common ratio, r = 8/2 = 4

Explicit formula:

t_n = arⁿ⁻¹

t_n = 2 × 4ⁿ⁻¹

Recursive formula:

t₁ = 2, t_n = 4 × t_n−1 for n ≥ 2

Now: 2 × 4ⁿ⁻¹ = 131072

⇒ 4ⁿ⁻¹ = 131072/2

⇒ 4ⁿ⁻¹ = 65536

Since: 65536 = 4⁸

So: n − 1 = 8

⇒ n = 9

Therefore, 131072 is the 9^th term.

Q9. The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the AP.

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

4^th term = a + 3d

8^th term = a + 7d

So, the sum of 4^th and 8^th term:

(a + 3d) + (a + 7d) = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12 …(1)

6^th term = a + 5d

10^th term = a + 9d

So, the sum of 6^th and 10^th term:

(a + 5d) + (a + 9d) = 44

⇒ 2a + 14d = 44

⇒ a + 7d = 22 …(2)

Subtracting (1) from (2), we get:

2d = 10

⇒ d = 5

Substituting d = 5 in (1), we have:

a + 5(5) = 12

⇒ a + 25 = 12

⇒ a = −13

First three terms: a, a + d, a + 2d

= −13, −8, −3

Therefore, the first three terms are −13, −8, −3.

Q10. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2^nd hour, 4^th hour and n^th hour?

Answer: Initial number of bacteria = 30

The bacteria double every hour.

At the end of 1^st hour = 30 × 2 = 60

At the end of 2^nd hour:

= 30 × 2²

= 30 × 4

= 120

At the end of 4^th hour:

= 30 × 2⁴

= 30 × 16

= 480

At the end of n^th hour = 30 × 2ⁿ

Therefore:

At the end of 2^nd hour = 120 bacteria

At the end of 4^th hour = 480 bacteria

At the end of n^th hour = 30 × 2ⁿ bacteria.

Q11. Find a GP for which the sum of the first two terms is −4 and the fifth term is 4 times the third term.

Answer: Let the first term of the GP be 'a' and common ratio be 'r'. So, the required GP: a, ar, ar², ar³, ar⁴, …

Given: Sum of first two terms = −4

⇒ a + ar = −4

⇒ a(1 + r) = −4 …(1)

Also, fifth term is 4 times the third term.

Fifth term = ar⁴

Third term = ar²

So, ar⁴ = 4ar²

⇒ r² = 4 [Dividing by ar²]

⇒ r = 2 or r = −2

Case 1: r = 2

From (1): a(1 + 2) = −4

⇒ 3a = −4

⇒ a = −4/3

Hence, the GP: −4/3, −8/3, −16/3, −32/3, −64/3, …

Case 2: r = −2

From (1): a(1 − 2) = −4

⇒ −a = −4

⇒ a = 4

Hence, the GP: 4, −8, 16, −32, 64, …

Therefore, possible GPs are −4/3, −8/3, −16/3, −32/3, … and 4, −8, 16, −32, …

Q12. How many multiples of 4 lie between 10 and 250?

Answer: The smallest multiple of 4 greater than 10 is 12.

The largest multiple of 4 less than 250 is 248.

So, the required AP: 12, 16, 20, …, 248

Here: First term, a = 12

Common difference, d = 4

Last term = 248

t_n = 248

⇒ 248 = 12 + (n − 1)4 [Using formula t_n = a + (n − 1)d]

⇒ 248 − 12 = 4(n − 1)

⇒ 236 = 4(n − 1)

⇒ 59 = n − 1

⇒ n = 60

Therefore, 60 multiples of 4 lie between 10 and 250.

Related Study Material on Extramarks

• Important Questions for Class 9 Maths Chapter 1 The Use of Coordinates
• Important Questions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials
• Important Questions for Class 9 Maths Chapter 3 The World of Numbers
• Important Questions for Class 9 Maths Chapter 4 Exploring Algebraic Identities
• Important Questions for Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round
• Important Questions for Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area
• Important Questions for Class 9 Maths Chapter 7 Introduction to Probability
• Important Questions for Class 9 Maths Chapter 9 Propositions
• Important Questions for Class 9 Maths Chapter 10 Data
• Important Questions for Class 9 Maths Chapter 11 Algorithms
• Important Questions for Class 9 Maths Chapter 12 Quadrilaterals
• Important Questions for Class 9 Maths Chapter 13 Two Variables, One Line
• Important Questions for Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume