NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions

A sequence is an ordered list of numbers in which each number is called a term. In Class 9 Maths Chapter 8, students learn how sequences follow rules, how terms can be found using explicit and recursive formulas, and how arithmetic and geometric progressions help predict patterns.

NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 8 cover Predicting What Comes Next: Exploring Sequences and Progressions, where students learn how ordered number patterns can be described using rules. The chapter introduces sequences, terms, explicit formulas, recursive formulas, arithmetic progressions and geometric progressions.

Growing squares, taxi fares, salary increments, marble arrangements, bouncing balls and Sierpiński fractals show how patterns can be predicted in CBSE 2026-27 Maths problems. The exercise-wise solutions help students find nth terms, check whether a number belongs to a sequence and solve AP and GP-based word problems.

NCERT Solutions for Class 9 Maths Chapter 8 PDF Download

The NCERT Solutions for Class 9 Maths Chapter 8 PDF helps students revise sequence and progression questions in a formula-wise manner for CBSE 2026-27. It is useful for practising nth term questions, AP and GP problems, recursive rules, natural number sums and fractal-based patterns.

Students can use the PDF to revise:

1. Sequences and terms
2. Finite and infinite sequences
3. Explicit formula for nth term
4. Recursive formula
5. Virahānka-Fibonacci sequence
6. Arithmetic progression
7. Common difference
8. Sum of first n natural numbers
9. Geometric progression
10. Common ratio
11. Sierpiński triangle
12. Sierpiński square carpet

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

Exercise	Main Focus	Solution Type
Exercise Set 8.1	Explicit and recursive rules	Finding terms and checking sequence membership
Exercise Set 8.2	Arithmetic progressions and sums	AP nth term, salary, divisibility and natural number sums
Exercise Set 8.3	Geometric progressions and fractals	GP nth term, bouncing ball and Sierpiński carpet
End-of-Chapter Exercises	Mixed sequence revision	AP, GP, consecutive sums and recursive rules

NCERT Solutions for Class 9 Maths Chapter 8 Exercise Set 8.1

Exercise Set 8.1 focuses on explicit rules, recursive rules and sequences where each term is generated using a formula.

Question 1. Find the first five terms of the sequence in which the nth term is given by the following.

(i) tₙ = 3n - 4

Answer:
Substitute n = 1, 2, 3, 4, 5.

t₁ = 3(1) - 4 = -1
t₂ = 3(2) - 4 = 2
t₃ = 3(3) - 4 = 5
t₄ = 3(4) - 4 = 8
t₅ = 3(5) - 4 = 11

Final answer:
-1, 2, 5, 8, 11

(ii) tₙ = 2 - 5n

Answer:
Substitute n = 1, 2, 3, 4, 5.

t₁ = 2 - 5 = -3
t₂ = 2 - 10 = -8
t₃ = 2 - 15 = -13
t₄ = 2 - 20 = -18
t₅ = 2 - 25 = -23

Final answer:
-3, -8, -13, -18, -23

(iii) tₙ = n² - 2n + 3

Answer:
Substitute n = 1, 2, 3, 4, 5.

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

Final answer:
2, 3, 6, 11, 18

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

Answer:
Given:

tₙ = 5n - 3

For n = 10:

t₁₀ = 5(10) - 3
= 50 - 3
= 47

For n = 15:

t₁₅ = 5(15) - 3
= 75 - 3
= 72

Final answer:
10th term = 47, 15th term = 72

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

Answer:
For 97:

5n - 3 = 97
5n = 100
n = 20

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

For 172:

5n - 3 = 172
5n = 175
n = 35

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

Final answer:
Both 97 and 172 are terms of the sequence.

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

Answer:
Given:

tₙ = 607

5n - 3 = 607
5n = 610
n = 122

Final answer:
607 is the 122nd term.

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

Answer:
Given:

t₁ = -5

Each next term is 3 more than the previous term.

t₁ = -5
t₂ = -5 + 3 = -2
t₃ = -2 + 3 = 1
t₄ = 1 + 3 = 4
t₅ = 4 + 3 = 7

First five terms are:

-5, -2, 1, 4, 7

This is an AP with first term -5 and common difference 3.

tₙ = -5 + (n - 1)3
tₙ = 3n - 8

Check if 52 is a term:

3n - 8 = 52
3n = 60
n = 20

Final answer:
First five terms are -5, -2, 1, 4, 7. Yes, 52 is the 20th term.

Question 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.

Answer:
Given:

T₁ = 1
T₂ = 2
T₃ = 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

Final answer:
T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81

NCERT Solutions for Class 9 Maths Chapter 8 Exercise Set 8.2

Exercise Set 8.2 focuses on arithmetic progressions, nth term of an AP and the sum of first n natural numbers.

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

Answer:
First term:

a = 3

Common difference:

d = 8 - 3 = 5

nth term of an AP:

tₙ = a + (n - 1)d

For n = 10:

t₁₀ = 3 + (10 - 1)5
= 3 + 45
= 48

For n = 26:

t₂₆ = 3 + (26 - 1)5
= 3 + 125
= 128

Final answer:
10th term = 48, 26th term = 128

Question 2. Which term of the AP 21, 18, 15, … is -81? Also, is 0 a term of this AP?

Answer:
First term:

a = 21

Common difference:

d = 18 - 21 = -3

nth term:

tₙ = 21 + (n - 1)(-3)
= 21 - 3n + 3
= 24 - 3n

For -81:

24 - 3n = -81
-3n = -105
n = 35

So, -81 is the 35th term.

Now check 0:

24 - 3n = 0
3n = 24
n = 8

Since n is a natural number, 0 is also a term.

Final answer:
-81 is the 35th term, and 0 is the 8th term.

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

Answer:
First term:

a = 11

Common difference:

d = 8 - 11 = -3

nth term:

tₙ = a + (n - 1)d
= 11 + (n - 1)(-3)
= 11 - 3n + 3
= 14 - 3n

Recursive rule:

t₁ = 11
tₙ = tₙ₋₁ - 3, for n ≥ 2

Final answer:
tₙ = 14 - 3n; recursive rule: t₁ = 11, tₙ = tₙ₋₁ - 3

Question 4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.

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

3rd term:

a + 2d = 12 …(1)

50th term:

a + 49d = 106 …(2)

Subtract (1) from (2):

47d = 94
d = 2

Substitute in (1):

a + 2(2) = 12
a + 4 = 12
a = 8

29th term:

t₂₉ = a + 28d
= 8 + 28(2)
= 8 + 56
= 64

Final answer:
29th term = 64

Question 5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

Answer:
The 2-digit numbers divisible by 3 are:

12, 15, 18, …, 99

This is an AP with:

a = 12
d = 3
last term = 99

Find n:

99 = 12 + (n - 1)3
87 = 3(n - 1)
n - 1 = 29
n = 30

So, there are 30 such numbers.

Sum:

S = n/2 × (first term + last term)

= 30/2 × (12 + 99)
= 15 × 111
= 1665

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

Question 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?

Answer:
Initial salary = ₹5,00,000
Annual increment = ₹20,000

Let the salary reach ₹7,00,000 after n years.

Increase needed:

₹7,00,000 - ₹5,00,000 = ₹2,00,000

Number of increments:

2,00,000 / 20,000 = 10

Final answer:
Harish’s income reached ₹7,00,000 after 10 years.

Question 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?

Answer:
Total marbles:

1 + 2 + 3 + … + 25

Sum of first n natural numbers:

Sₙ = n(n + 1)/2

For n = 25:

S₂₅ = 25(26)/2
= 25 × 13
= 325

Final answer:
325 marbles

NCERT Solutions for Class 9 Maths Chapter 8 Exercise Set 8.3

Exercise Set 8.3 focuses on geometric progressions, common ratio, nth term of a GP, bouncing ball problems and fractal patterns.

Question 1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.

Answer:
Given:

Common ratio r = 2
8th term = 192

In a GP:

t₁₂ = t₈ × r⁴

Because the 12th term is 4 places after the 8th term.

t₁₂ = 192 × 2⁴
= 192 × 16
= 3072

Final answer:
12th term = 3072

Question 2. Find the 10th and nth terms of the GP: 5, 25, 125, …

Answer:
First term:

a = 5

Common ratio:

r = 25/5 = 5

nth term of a GP:

tₙ = arⁿ⁻¹

So:

tₙ = 5 × 5ⁿ⁻¹
= 5ⁿ

10th term:

t₁₀ = 5¹⁰
= 9765625

Final answer:
tₙ = 5ⁿ and t₁₀ = 9765625

Question 3. A sequence is given by the recursive rule t₁ = 2, tₙ₊₁ = 3tₙ - 2 for n ≥ 1. Which term of the sequence is 730?

Answer:
Find the first few terms:

t₁ = 2
t₂ = 3(2) - 2 = 4
t₃ = 3(4) - 2 = 10
t₄ = 3(10) - 2 = 28
t₅ = 3(28) - 2 = 82
t₆ = 3(82) - 2 = 244
t₇ = 3(244) - 2 = 730

Final answer:
730 is the 7th term.

Question 4. Which term of the GP 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.

Answer:
First term:

a = 2

Common ratio:

r = 6/2 = 3

Explicit formula:

tₙ = 2 × 3ⁿ⁻¹

Now find n:

2 × 3ⁿ⁻¹ = 4374

3ⁿ⁻¹ = 4374/2
= 2187

2187 = 3⁷

So:

n - 1 = 7
n = 8

Recursive formula:

t₁ = 2
tₙ = 3tₙ₋₁, for n ≥ 2

Final answer:
4374 is the 8th term. Explicit formula: tₙ = 2 × 3ⁿ⁻¹. Recursive formula: t₁ = 2, tₙ = 3tₙ₋₁.

Question 5. A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell.

(i) What height does the ball reach after the 5th bounce?

Answer:
Initial height = 80 m

Each bounce reaches 60% = 0.6 of the previous height.

After 1st bounce:

80 × 0.6 = 48 m

After 5th bounce:

80 × (0.6)⁵
= 80 × 0.07776
= 6.2208 m

Final answer:
6.2208 m

(ii) What is the total vertical distance travelled by the time it hits the ground for the 6th time?

Answer:
The ball first falls 80 m.

Then it rises and falls after each bounce.

Heights reached after bounces:

1st bounce = 48 m
2nd bounce = 28.8 m
3rd bounce = 17.28 m
4th bounce = 10.368 m
5th bounce = 6.2208 m

By the time it hits the ground for the 6th time, it has:

1. Fallen initially from 80 m.
2. Gone up and down for the first five bounce heights.

Total distance:

= 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208)

= 80 + 2(110.6688)
= 80 + 221.3376
= 301.3376 m

Final answer:
301.3376 m

Question 6. Which term of the sequence 2√2, 2, 2√2/2, … is 1/128?

Answer:
The given sequence is a GP.

First term:

a = 2√2

Common ratio:

r = 1/√2

nth term:

tₙ = 2√2 × (1/√2)ⁿ⁻¹

Write in powers of 2:

2√2 = 2³ᐟ²
1/√2 = 2⁻¹ᐟ²

So:

tₙ = 2³ᐟ² × 2⁻(n-1)/2
= 2(3 - n + 1)/2
= 2(4 - n)/2
= 2²⁻ⁿᐟ²

Set:

2²⁻ⁿᐟ² = 1/128 = 2⁻⁷

So:

2 - n/2 = -7
-n/2 = -9
n = 18

Final answer:
1/128 is the 18th term.

Question 7. Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet.

(i) How many red squares are there in Stages 0 to 3?

Answer:
Stage 0 has 1 red square.

At each stage, every red square is divided into 9 smaller squares and the centre square is removed. So each red square gives 8 red squares in the next stage.

Stage 0 = 1
Stage 1 = 8
Stage 2 = 8² = 64
Stage 3 = 8³ = 512

Final answer:
1, 8, 64, 512

(ii) Predict the number of red squares in Stages 4 and 5.

Answer:
Stage 4 = 8⁴ = 4096
Stage 5 = 8⁵ = 32768

Final answer:
Stage 4 = 4096, Stage 5 = 32768

(iii) Find a rule for the number of red squares at the nth stage. Write the explicit formula and recursive formula.

Answer:
At Stage n, the number of red squares is:

tₙ = 8ⁿ

This is the explicit formula if counting starts from Stage 0.

Recursive formula:

t₀ = 1
tₙ = 8tₙ₋₁, for n ≥ 1

Final answer:
Explicit formula: tₙ = 8ⁿ. Recursive formula: t₀ = 1, tₙ = 8tₙ₋₁.

(iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area in Stages 4 and 5? Find the explicit and recursive formula.

Answer:
At each stage, 8 out of 9 equal parts remain.

So, the area is multiplied by 8/9 at each stage.

Stage 0 area = 1
Stage 1 area = 8/9
Stage 2 area = (8/9)²
Stage 3 area = (8/9)³
Stage 4 area = (8/9)⁴
Stage 5 area = (8/9)⁵

Explicit formula:

sₙ = (8/9)ⁿ

Recursive formula:

s₀ = 1
sₙ = (8/9)sₙ₋₁, for n ≥ 1

As n increases, the area keeps decreasing and gets closer to 0.

Final answer:
sₙ = (8/9)ⁿ; recursive rule: s₀ = 1, sₙ = (8/9)sₙ₋₁.

NCERT Solutions for Class 9 Maths Chapter 8 End-of-Chapter Exercises

The end-of-chapter exercises include AP, GP, consecutive natural number sums, bacteria growth, recursive rules and sequence-based reasoning.

Question 1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

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

11th term:

a + 10d = 38 …(1)

16th term:

a + 15d = 73 …(2)

Subtract (1) from (2):

5d = 35
d = 7

Substitute in (1):

a + 10(7) = 38
a + 70 = 38
a = -32

31st term:

t₃₁ = a + 30d
= -32 + 30(7)
= -32 + 210
= 178

Final answer:
31st term = 178

Question 2. Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

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

Third term:

a + 2d = 16 …(1)

7th term exceeds 5th term by 12:

t₇ - t₅ = 12

(a + 6d) - (a + 4d) = 12
2d = 12
d = 6

Substitute in (1):

a + 2(6) = 16
a + 12 = 16
a = 4

Therefore, the AP is:

4, 10, 16, 22, 28, …

Final answer:
AP = 4, 10, 16, 22, 28, …

Question 3. How many three-digit numbers are divisible by 7?

Answer:
Smallest three-digit number divisible by 7:

105

Largest three-digit number divisible by 7:

994

The numbers form an AP:

105, 112, 119, …, 994

Here:

a = 105
d = 7
last term = 994

Find n:

994 = 105 + (n - 1)7
889 = 7(n - 1)
n - 1 = 127
n = 128

Final answer:
128 three-digit numbers are divisible by 7.

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

Answer:
Smallest multiple of 4 greater than 10 is 12.

Largest multiple of 4 less than 250 is 248.

The multiples are:

12, 16, 20, …, 248

Here:

a = 12
d = 4
last term = 248

Find n:

248 = 12 + (n - 1)4
236 = 4(n - 1)
n - 1 = 59
n = 60

Final answer:
60 multiples of 4 lie between 10 and 250.

Question 5. 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 GP be:

a, ar, ar², ar³, ar⁴, …

Given:

a + ar = -4

So:

a(1 + r) = -4 …(1)

Fifth term is 4 times third term:

ar⁴ = 4ar²

Assuming a ≠ 0:

r² = 4
r = 2 or r = -2

If r = 2:

a(1 + 2) = -4
3a = -4
a = -4/3

GP is:

-4/3, -8/3, -16/3, …

If r = -2:

a(1 - 2) = -4
-a = -4
a = 4

GP is:

4, -8, 16, -32, 64, …

Final answer:
Possible GPs are -4/3, -8/3, -16/3, … and 4, -8, 16, -32, …

Question 6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Answer:
Let 100 be written as the sum of k consecutive natural numbers starting from a.

100 = a + (a + 1) + … + (a + k - 1)

100 = k/2[2a + k - 1]

Check possible values of k.

One-term sum:

100

Two-term sum:

49 + 51 is not consecutive because 50 + 51 = 101 and 49 + 50 = 99. So no 2-term expression.

Four-term sum:

22 + 23 + 24 + 25 = 94
23 + 24 + 25 + 26 = 98
24 + 25 + 26 + 27 = 102
So no 4-term expression.

Five-term sum:

18 + 19 + 20 + 21 + 22 = 100

Eight-term sum:

9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100

Final answer:
100 = 100 = 18 + 19 + 20 + 21 + 22 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

Question 7. The number of bacteria in a culture doubles every hour. If there were 30 bacteria originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?

Answer:
Initial number = 30

The number doubles every hour.

After 1 hour:

30 × 2 = 60

After 2 hours:

30 × 2² = 120

After 4 hours:

30 × 2⁴ = 480

After n hours:

30 × 2ⁿ

Final answer:
After 2 hours = 120, after 4 hours = 480, after nth hour = 30 × 2ⁿ

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

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

4th term:

a + 3d

8th term:

a + 7d

Given:

(a + 3d) + (a + 7d) = 24
2a + 10d = 24
a + 5d = 12 …(1)

6th term:

a + 5d

10th term:

a + 9d

Given:

(a + 5d) + (a + 9d) = 44
2a + 14d = 44
a + 7d = 22 …(2)

Subtract (1) from (2):

2d = 10
d = 5

Substitute in (1):

a + 5(5) = 12
a + 25 = 12
a = -13

First three terms:

a = -13
a + d = -8
a + 2d = -3

Final answer:
-13, -8, -3

Question 9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1000.

Answer:
Sum of first n natural numbers:

Sₙ = n(n + 1)/2

We need:

n(n + 1)/2 > 1000

n(n + 1) > 2000

Try n = 44:

44 × 45 = 1980
S₄₄ = 990

Try n = 45:

45 × 46 = 2070
S₄₅ = 1035

So, the smallest n is 45.

Final answer:
n = 45

Question 10. Which term of the GP 2, 8, 32, … is 131072? Write the explicit formula and recursive formula.

Answer:
First term:

a = 2

Common ratio:

r = 8/2 = 4

Explicit formula:

tₙ = 2 × 4ⁿ⁻¹

Find n:

2 × 4ⁿ⁻¹ = 131072

4ⁿ⁻¹ = 65536

Since:

65536 = 4⁸

n - 1 = 8
n = 9

Recursive formula:

t₁ = 2
tₙ = 4tₙ₋₁, for n ≥ 2

Final answer:
131072 is the 9th term. Explicit formula: tₙ = 2 × 4ⁿ⁻¹. Recursive formula: t₁ = 2, tₙ = 4tₙ₋₁.

Question 11. The sum of the first three terms of a GP is 13/12 and their product is -1. Find the common ratio and the terms.

Answer:
Let the three terms of the GP be:

a/r, a, ar

Their product is:

(a/r) × a × ar = a³

Given product = -1

So:

a³ = -1
a = -1

The terms are:

-1/r, -1, -r

Their sum is 13/12:

-1/r - 1 - r = 13/12

Multiply by 12r:

-12 - 12r - 12r² = 13r

12r² + 25r + 12 = 0

Factor:

12r² + 25r + 12 = (3r + 4)(4r + 3)

So:

r = -4/3 or r = -3/4

If r = -4/3, terms are:

3/4, -1, 4/3

If r = -3/4, terms are:

4/3, -1, 3/4

Final answer:
Common ratio = -4/3 or -3/4. Terms are 3/4, -1, 4/3 or 4/3, -1, 3/4.

Question 12. If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

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

4th term:

x = ar³

10th term:

y = ar⁹

16th term:

z = ar¹⁵

Now:

y/x = ar⁹/ar³ = r⁶

z/y = ar¹⁵/ar⁹ = r⁶

Since:

y/x = z/y

Therefore, x, y and z are in GP.

Final answer:
x, y, z are in GP because their consecutive ratios are equal.

Question 13. 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/r, a, ar

Their sum:

a/r + a + ar = 26 …(1)

Sum of squares:

a²/r² + a² + a²r² = 364 …(2)

Let:

x = a/r, y = a, z = ar

Then x, y, z are in GP and:

x + y + z = 26
x² + y² + z² = 364

Also:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

26² = 364 + 2(xy + yz + zx)

676 - 364 = 2(xy + yz + zx)

312 = 2(xy + yz + zx)

xy + yz + zx = 156

Checking simple GP triples with sum 26 and square sum 364 gives:

2, 6, 18

Sum = 2 + 6 + 18 = 26
Squares = 4 + 36 + 324 = 364

Final answer:
The terms are 2, 6 and 18.

Question 14. Suppose P₁ = 1, P₂ = 2 and for n > 2, Pₙ = P₁ + P₂ + … + Pₙ₋₁ + 1. Find P₁ to P₈. Can you find a simpler recursive formula and an explicit formula?

Answer:
Given:

P₁ = 1
P₂ = 2

For n > 2:

Pₙ = P₁ + P₂ + … + Pₙ₋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

So the sequence is:

P₁ = 1, P₂ = 2, P₃ = 4, P₄ = 8, P₅ = 16, P₆ = 32, P₇ = 64, P₈ = 128

Simpler recursive formula:

P₁ = 1
Pₙ = 2Pₙ₋₁, for n ≥ 2

Explicit formula:

Pₙ = 2ⁿ⁻¹

Final answer:
1, 2, 4, 8, 16, 32, 64, 128; Pₙ = 2Pₙ₋₁; Pₙ = 2ⁿ⁻¹

Question 15. Suppose W₁ = 1, W₂ = 2 and for n > 2, Wₙ = W₁ + W₂ + … + Wₙ₋₂ + 2. Find W₁ to W₈. Do you recognise this sequence?

Answer:
Given:

W₁ = 1
W₂ = 2

For n > 2:

Wₙ = W₁ + W₂ + … + Wₙ₋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

Final answer:
W₁ to W₈ are 1, 2, 3, 5, 8, 13, 21, 34. This is the Virahānka-Fibonacci sequence.

Topics Covered in NCERT Solutions for Class 9 Maths Chapter 8

1. Sequences
2. Terms of a sequence
3. Finite sequences
4. Infinite sequences
5. Natural number sequence
6. Odd number sequence
7. Triangular numbers
8. Square numbers
9. Explicit formula
10. Recursive formula
11. Virahānka-Fibonacci sequence
12. Arithmetic progression
13. Common difference
14. nth term of an AP
15. Visualising an AP
16. Sum of first n natural numbers
17. Geometric progression
18. Common ratio
19. nth term of a GP
20. Fractals
21. Sierpiński triangle
22. Sierpiński square carpet

Important Formulas in NCERT Solutions for Class 9 Maths Chapter 8

Concept	Formula / Rule
nth term of odd numbers	tₙ = 2n - 1
nth triangular number	tₙ = n(n + 1)/2
Sum of first n natural numbers	Sₙ = n(n + 1)/2
nth term of AP	tₙ = a + (n - 1)d
Recursive rule for AP	t₁ = a, tₙ = tₙ₋₁ + d
General AP	a, a + d, a + 2d, …
nth term of GP	tₙ = arⁿ⁻¹
Recursive rule for GP	t₁ = a, tₙ = rtₙ₋₁
General GP	a, ar, ar², ar³, …
Sierpiński triangle black triangles	tₙ = 3ⁿ
Sierpiński triangle area	sₙ = (3/4)ⁿ
Sierpiński square carpet red squares	tₙ = 8ⁿ
Sierpiński square carpet area	sₙ = (8/9)ⁿ

NCERT Class 9 Maths Ganita Manjari 2026 Chapter Solutions

Chapter	Title
Chapter 1	Orienting Yourself: The Use of Coordinates
Chapter 2	Introduction to Linear Polynomials
Chapter 3	The World of Numbers
Chapter 4	Exploring Algebraic Identities
Chapter 5	I’m Up and Down, and Round and Round
Chapter 6	Measuring Space: Perimeter and Area
Chapter 7	The Mathematics of Maybe: Introduction to Probability
Chapter 8	Predicting What Comes Next: Exploring Sequences and Progressions