NCERT Class 9 Maths Ganita Manjari Chapter 8 Solutions (2026-27 New Syllabus)

Class 9 Maths Ganita Manjari Chapter 8 Exercise Set 8.1 Solutions

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.

Answer:
(i) tₙ = 3n − 4
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
So, the first five terms: −1, 2, 5, 8, 11

(ii) tₙ = 2 − 5n
t₁ = 2 − 5(1) = −3
t₂ = 2 − 5(2) = −8
t₃ = 2 − 5(3) = −13
t₄ = 2 − 5(4) = −18
t₅ = 2 − 5(5) = −23
So, the first five terms: −3, −8, −13, −18, −23

(iii) 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
So, the first five terms: 2, 3, 6, 11, 18.

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

Answer:
Given: tₙ = 5n − 3
10th term:
t₁₀ = 5(10) − 3
= 50 − 3
= 47

15th term:
t₁₅ = 5(15) − 3
= 75 − 3
= 72

Therefore:
10th term = 47
15th term = 72.

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

Answer:
Given: tₙ = 5n − 3
For 97:
Let tₙ = 97
⇒ 5n − 3 = 97
⇒ 5n = 100
⇒ n = 20
Since n = 20 is a natural number, 97 is a term of the sequence.
So, 97 is the 20th term.

For 172:
Let tₙ = 172
⇒ 5n − 3 = 172
⇒ 5n = 175
⇒ n = 35
Since n = 35 is a natural number, 172 is also a term of the sequence.
So, 172 is the 35th term.

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

Answer:
Given: tₙ = 5n − 3
Let tₙ = 607
⇒ 5n − 3 = 607
⇒ 5n = 610
⇒ n = 122
Therefore, 607 is the 122nd term of the 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?

Answer:
Given: t₁ = −5 and tₙ₊₁ = tₙ + 3
First five terms:
t₁ = −5
t₂ = t₁ + 3 = −5 + 3 = −2 [Using tₙ₊₁ = tₙ + 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
This is an arithmetic sequence with first term a = −5 common difference d = 3.

Formula: tₙ = a + (n − 1)d
tₙ = −5 + (n − 1)3
= −5 + 3n − 3
= 3n − 8

Now checking whether 52 is a term:
Let tₙ = 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 20th term.

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
Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ (for n ≥ 4)

Now computing step by step:
T₄ = T₃ + T₂ + T₁ [Using Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ and putting n = 4]
= 4 + 2 + 1
= 7

T₅ = T₄ + T₃ + T₂ [Using Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ and putting n = 5]
= 7 + 4 + 2
= 13

T₆ = T₅ + T₄ + T₃ [Using Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ and putting n = 6]
= 13 + 7 + 4
= 24

T₇ = T₆ + T₅ + T₄ [Using Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ and putting n = 7]
= 24 + 13 + 7
= 44

T₈ = T₇ + T₆ + T₅ [Using Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ and putting n = 8]
= 44 + 24 + 13
= 81.

Class 9 Maths Ganita Manjari Chapter 8 Exercise Set 8.2 Solutions

Exercise Set 8.2

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

Answer:
Given AP: 3, 8, 13, 18, …
First term, a = 3
Common difference, d = 8 − 3 = 5
Formula: tₙ = a + (n − 1)d

10th term: t₁₀ = 3 + (10 − 1)5
= 3 + 45
= 48

26th term: t₂₆ = 3 + (26 − 1)5
= 3 + 125
= 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.

Answer:
Given AP: 21, 18, 15, …
First term, a = 21
Common difference, d = 18 − 21 = −3
Formula: tₙ = a + (n − 1)d

So, tₙ = 21 + (n − 1)(−3) [Putting the values of a and d]
= 21 − 3n + 3
= 24 − 3n

Now, checking for −81:
Let tₙ = 81
⇒ 24 − 3n = −81
⇒ −3n = −105
⇒ n = 35
Therefore, −81 is the 35th term.

Now, checking whether 0 is a term:
Let tₙ = 0
⇒ 24 − 3n = 0
⇒ 3n = 24
⇒ n = 8
Since n = 8 is a natural number, 0 is a term of this AP.
Therefore, 0 is the 8th term.

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

Answer:
Given AP: 11, 8, 5, 2, …
First term, a = 11
Common difference, d = 8 − 11 = −3
Formula: tₙ = a + (n − 1)d

tₙ = 11 + (n − 1)(−3) [Putting the values of a and d]
= 11 − 3n + 3
= 14 − 3n
Therefore, nth term: tₙ = 14 − 3n

Recursive rule:
t₁ = 11
tₙ = 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’ 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’.)
Answer:
Let the first term be a and common difference be d.
Given: 3rd term = 12
So, a + 2d = 12 …(1)
The AP has 50 terms and last term = 106.
So, 50th term = 106
⇒ a + 49d = 106 …(2)
Subtracting equation (1) from equation (2), we get:
(a + 49d) − (a + 2d) = 106 − 12
⇒ 47d = 94
⇒ d = 2

Now, Substituting d = 2 in equation (1), we have:
a + 2(2) = 12
⇒ a + 4 = 12
⇒ a = 8

Find the 29th term:
t₂₉ = a + (29 − 1)d
= 8 + 28 × 2
= 8 + 56
= 64
Therefore, the 29th term is 64.

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
We can observe that this is an AP where:
First term, a = 12
Common difference, d = 15 – 12 = 3
Last term = 99

Using the formula: tₙ = a + (n − 1)d, we have
99 = 12 + (n − 1)3
⇒ 99 − 12 = 3(n − 1)
⇒ 87 = 3(n − 1)
⇒ 29 = n − 1
⇒ n = 30
So, there are 30 two-digit numbers divisible by 3.

Now, sum of these numbers: Sₙ = n/2 × (first term + last term)
⇒ S₃₀ = 30/2 × (12 + 99)
= 15 × 111
= 1665
Therefore:
Number of 2-digit numbers divisible by 3 = 30
Sum of all these numbers = 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?

Answer:
Initial salary = ₹5,00,000
Annual increment = ₹20,000
This forms an AP: ₹5,00,000, ₹5,20,000, ₹5,40,000, …
We have to find when salary becomes ₹7,00,000.

Using the formula: tₙ = a + (n − 1)d
⇒ ₹7,00,000 = ₹5,00,000 + (n − 1)₹20,000
⇒ ₹2,00,000 = (n − 1)₹20,000
⇒ n − 1 = 10
⇒ n = 11
Therefore, Harish’s income reached ₹7,00,000 in the 11th year.
So, it took 10 increments, i.e., after 10 years of work.

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:
Number of marbles: 1 + 2 + 3 + … + 25
Formula: Sₙ = n(n + 1)/2
⇒ S₂₅ = 25(25 + 1)/2 [Here, n = 25]
= 25 × 26 / 2
= 25 × 13
= 325
Therefore, the child uses 325 marbles in all.

Class 9 Maths Ganita Manjari Chapter 8 Exercise Set 8.3 Solutions

Exercise Set 8.3

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
So, t₈ = ar⁷ = 192 [Using tₙ = arⁿ⁻¹]
⇒ a(2)⁷ = 192 [Since r =2]
⇒ a × 128 = 192
⇒ a = 192/128 = 3/2
⇒ a = 3/2

Now, t₁₂ = ar¹¹ [Using tₙ = arⁿ⁻¹]
= 3/2 × (2)¹¹
= 3 × 2¹⁰
= 3 × 1024
= 3072
Therefore, the 12th term is 3072.

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

Answer:
Given GP: 5, 25, 125, …
First term, a = 5
Common ratio, r = a₂/a₁ = 25/5 = 5
So, tₙ = 5 × 5ⁿ⁻¹ [Using Formula tₙ = arⁿ⁻¹]
= 5ⁿ
Therefore: nth term = 5ⁿ
Hence, 10th term: t₁₀ = 5¹⁰ = 9765625.

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:
Given: t₁ = 2
tₙ₊₁ = 3tₙ − 2

Finding terms step by step:
t₁ = 2
t₂ = 3t₁ − 2 [Using tₙ₊₁ = 3tₙ − 2]
= 3(2) − 2
= 4

Similarly, t₃ = 3t₂ − 2
= 3(4) − 2
= 10

t₄ = 3t₃ − 2
= 3(10) − 2
= 28

t₅ = 3t₄ − 2
= 3(28) − 2
= 82

t₆ = 3t₅ − 2
= 3(82) − 2
= 244

t₇ = 3t₆ − 2
= 3(244) − 2
= 730

Therefore, 730 is the 7th term of the sequence.

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:
Given GP: 2, 6, 18, …
First term, a = 2
Common ratio, r = a₂/a₁ = 6/2 = 3

Explicit formula: tₙ = arⁿ⁻¹
⇒ tₙ = 2 × 3ⁿ⁻¹ [As a = 2 and r = 3]

Let tₙ = 4374
⇒ 2 × 3ⁿ⁻¹ = 4374
⇒ 3ⁿ⁻¹ = 4374/2
⇒ 3ⁿ⁻¹ = 2187
⇒ 3ⁿ⁻¹ = 3⁷ [Since 2187 = 3⁷]
⇒ n − 1 = 7
⇒ n = 8
Therefore, 4374 is the 8th term.
Recursive formula:
t₁ = 2
tₙ = 3tₙ₋₁, for n ≥ 2.

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. It continues bouncing in this way—each time rising to 60% of the previous height.

(i) What height does the ball reach after the 5th bounce?
Answer:
Initial height = 80 m
Bounce ratio = 60% = 0.6
Height after 1st bounce:
= 80 × 0.6
= 48 m

Height after 2nd bounce:
= 48 × 0.6
= 28.8 m

Height after 3rd bounce:
= 28.8 × 0.6
= 17.28 m

Height after 4th bounce:
= 17.28 × 0.6
= 10.368 m

Height after 5th bounce:
= 10.368 × 0.6
= 6.2208 m
Therefore, the ball reaches 6.2208 m after the 5th bounce.

(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
Answer:
The ball first falls 80 m to hit the ground for the 1st time.
Then it rises and falls after each bounce.
Heights 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
To hit the ground for the 6th time, it travels the total distance
= 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208)
= 80 + 2(110.6688)
= 80 + 221.3376
= 301.3376 m
Therefore, the total vertical distance travelled is 301.3376 m.

6. Which term of the sequence 2, 2√2, 4, … is 128?

Answer:
Given sequence: 2, 2√2, 4, …
This is a GP.
First term, a = 2
Common ratio: r = a₂/a₁ = 2√2/2 = √2
So, tₙ = 2(√2)ⁿ⁻¹ [Using the formula tₙ = arⁿ⁻¹]
Let tₙ = 128
⇒ 2(√2)ⁿ⁻¹ = 128
⇒ (√2)ⁿ⁻¹ = 64
⇒ (2¹/²)ⁿ⁻¹ = 2⁶ [Since 64 = 2⁶ and √2 = 2¹/²]
⇒ 2^(n−1)/2 = 2⁶
⇒ (n − 1)/2 = 6 [Comparing the powers of 2]
⇒ n − 1 = 12
⇒ n = 13
Therefore, 128 is the 13th term.

7. Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on.

Look at Fig. 8.12 and try to answer the following questions.
(i) How many red squares are there in Stages 0 to 3?
(ii) Can you predict the number of red squares in Stages 4 and 5?
(iii) Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.
(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 of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?
Answer:
(i) Stage 0: 1 red square
Stage 1: 8 red squares
Stage 2: 8² = 64 red squares
Stage 3: 8³ = 512 red squares
Therefore: Stages 0 to 3 have 1, 8, 64 and 512 red squares respectively.

(ii) The number of red squares is multiplied by 8 at each stage.
Stage 4 = 8⁴ = 4096
Stage 5 = 8⁵ = 32768
Therefore:
Stage 4 has 4096 red squares.
Stage 5 has 32768 red squares.

(iii) At each stage, every red square is replaced by 8 smaller red squares.
So, the number of red squares forms the sequence:
1, 8, 64, 512, …
Explicit formula: tₙ = 8ⁿ
Here, Stage 0 corresponds to n = 0.
Recursive formula:
t₀ = 1
tₙ = 8tₙ₋₁, for n ≥ 1

(iv) At each stage, the square is divided into 9 equal parts and the centre part is removed.
So, the remaining red area becomes 8/9 of the previous area.
Stage 0 area = 1 square unit
Stage 1 area = 8/9
Stage 2 area = (8/9)² = 64/81
Stage 3 area = (8/9)³ = 512/729
Stage 4 area = (8/9)⁴ = 4096/6561
Stage 5 area = (8/9)⁵ = 32768/59049

Explicit formula Aₙ = (8/9)ⁿ
Here, Stage 0 corresponds to n = 0.
Recursive formula:
A₀ = 1
Aₙ = (8/9)Aₙ₋₁, for n ≥ 1
As n increases, the area keeps decreasing and gets closer and closer to 0, but it never becomes exactly 0.

Class 9 Maths Ganita Manjari Chapter 8 End-of-Chapter Exercises Solutions

End-of-Chapter Exercises

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

Answer:
Let first term = a and common difference = d

Given:
11th term ⇒ a + 10d = 38 …(1)
16th term ⇒ a + 15d = 73 …(2)

Subtracting (1) from (2), we get:
(a + 15d) − (a + 10d) = 73 − 38
⇒ 5d = 35
⇒ d = 7

Substituting d = 7 in (1), we have:
a + 10(7) = 38
⇒ a + 70 = 38
⇒ a = -32

Now, 31st term: t₃₁ = a + 30d
= -32 + 30(7)
= -32 + 210
= 178
Therefore, the 31st term is 178.

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

Answer:
Let first term = a and common difference = d
Third term: a + 2d = 16 …(1)
According to question:
7th term – 5th term = 12 [Since 7th term exceeds 5th term by 12]
⇒ (a + 6d) − (a + 4d) = 12
⇒ 2d = 12
⇒ d = 6

Substituting d = 6 in (1), we get:
a + 2(6) = 16
⇒ a + 12 = 16
⇒ a = 4

Thus, AP: 4, 10, 16, 22, 28, …
Therefore: First term = 4 and common difference = 6.

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

(Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)
Answer:
The smallest three-digit number divisible by 7 is 105.
The largest three-digit number divisible by 7 is 994.
So, the required AP: 105, 112, 119, …, 994

Here: First term, a = 105
Common difference, d = 7
Last term = 994

Let tₙ = 994
⇒ 994 = 105 + (n − 1)7 [Using Formula tₙ = a + (n − 1)d]
⇒ 994 − 105 = 7(n − 1)
⇒ 889 = 7(n − 1)
⇒ 127 = n − 1
⇒ n = 128
Therefore, 128 three-digit numbers are divisible by 7.

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

(Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 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

Let tₙ = 994
⇒ 248 = 12 + (n − 1)4 [Using Formula tₙ = 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.

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 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, …

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

Answer:
We need consecutive natural numbers whose sum is 100.
Possible lengths:

1 term: 100
So, 100 = 100

5 terms:
Let the numbers be: x − 2, x − 1, x, x + 1, x + 2
Sum = 5x
⇒ 5x = 100
⇒ x = 20
So, 100 = 18 + 19 + 20 + 21 + 22

8 terms:
Let the numbers be: x, x + 1, x + 2, …, x + 7
Sum = 8x + 28
⇒ 8x + 28 = 100
⇒ 8x = 72
⇒ x = 9
So, 100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

Therefore, all possible ways are:
100 = 100
100 = 18 + 19 + 20 + 21 + 22
100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16.

7. 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 2nd hour, 4th hour and nth hour?

Answer:
Initial number of bacteria = 30
The bacteria double every hour.
At the end of 1st hour = 30 × 2 = 60
At the end of 2nd hour:
= 30 × 2²
= 30 × 4
= 120

At the end of 4th hour:
= 30 × 2⁴
= 30 × 16
= 480

At the end of nth hour = 30 × 2ⁿ

Therefore:
At the end of 2nd hour = 120 bacteria
At the end of 4th hour = 480 bacteria
At the end of nth hour = 30 × 2ⁿ bacteria.

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 of the AP.

Answer:
Let the first term be a and common difference be d.
4th term = a + 3d
8th term = a + 7d
So, the sum of 4th and 8th term:
(a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 …(1)

6th term = a + 5d
10th term = a + 9d
So, the sum of 6th and 10th 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.

9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Answer:
Sum of first n natural numbers:
Sₙ = n(n + 1)/2

We need: n(n + 1)/2 > 1000
⇒ n(n + 1) > 2000

Now checking:
44 × 45 = 1980, so S₄₄ = 990
45 × 46 = 2070, so S₄₅ = 1035
Since 1035 > 1000,
Therefore, the smallest value of n is 45.

10. Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.

Answer:
Given GP: 2, 8, 32, …
First term, a = 2
Common ratio, r = 8/2 = 4

Explicit formula:
tₙ = arⁿ⁻¹
tₙ = 2 × 4ⁿ⁻¹

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 9th term.

Recursive formula:
t₁ = 2
tₙ = 4tₙ₋₁, for n ≥ 2.

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
Product: (a/r) × a × ar = a³
According to question, product = -1
⇒ a³ = -1
⇒ a = -1
So, the three terms are -1/r, -1, -r
Sum: -1/r – 1 – r = 13/12
⇒ -12 – 12r – 12r² = 13r
⇒ 12r² + 25r + 12 = 0
⇒ 12r² + 16r + 9r + 12 = 0
⇒ 4r(3r + 4) + 3(3r + 4) = 0
⇒ (4r + 3)(3r + 4) = 0
So: r = -3/4 or r = -4/3

If r = -3/4:
Terms are 4/3, -1, 3/4
If r = -4/3:
Terms are 3/4, -1, 4/3
Therefore, the common ratio is -3/4 or -4/3, and the terms are 4/3, -1, 3/4 or 3/4, -1, 4/3.

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 first term of the GP be a and common ratio be r.
Then:
4th term = ar³ = x
10th term = ar⁹ = y
16th 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.

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, 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.

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

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
Pₙ = 2Pₙ₋₁, for n ≥ 2

Explicit formula Pₙ = 2ⁿ⁻¹.

15. Suppose W₁ = 1, W₂ = 2 and for n > 2, Wₙ = W₁ + W₂ + … + Wₙ₋₂ + 2. Find the values of W₁, W₂, …, 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

Therefore: W₁, W₂, …, W₈ are 1, 2, 3, 5, 8, 13, 21, 34
Yes, this is the Virahanka-Fibonacci sequence.

Important Formulas of Class 9 Maths Ganita Manjari Chapter 8

These are the formulas every student must know before sitting any exam on Sequences and Progressions. Each one should be memorised with its meaning, not just its symbols.

Formula	What It Means
tₙ = a + (n − 1)d	nth term of an Arithmetic Progression
t₁ = a, tₙ = tₙ₋₁ + d for n ≥ 2	Recursive rule for an AP
tₙ = arⁿ⁻¹	nth term of a Geometric Progression
t₁ = a, tₙ = r × tₙ₋₁ for n ≥ 2	Recursive rule for a GP
Sₙ = n(n + 1) / 2	Sum of the first n natural numbers
tₙ = n(n + 1) / 2	nth triangular number
d = t₂ − t₁ = t₃ − t₂	Common difference of an AP
r = t₂/t₁ = t₃/t₂	Common ratio of a GP

What each variable means:

a = first term of the AP or GP
d = common difference (AP only)
r = common ratio (GP only)
n = position number of the term (always a natural number)
tₙ = value of the term at position n
Sₙ = sum of the first n natural numbers

Three things students always confuse:

In tₙ = a + (n − 1)d, the exponent on (n − 1) is 1, not n. Do not write tₙ = a + nd.
In tₙ = arⁿ⁻¹, the exponent is n − 1, not n. When n = 1, you must get t₁ = a × r⁰ = a.
d is found by subtracting a term from the term after it (t₂ − t₁), not the term before it.

Summary of Chapter 8 Ganita Manjari – Sequences and Progressions

What is a Sequence?

A sequence is an ordered list of numbers where each number is called a term. Terms are labelled using subscripts – t₁ for the first, t₂ for the second and tₙ for the nth term. Sequences can be finite (fixed number of terms) or infinite (continuing indefinitely, shown by …). The terms can be positive, negative, fractions or any real number. Some sequences follow a clear rule; others, like the prime numbers, do not have a simple pattern.

Explicit Rule vs Recursive Rule

An explicit rule expresses tₙ directly in terms of n — you can find any term without knowing others. For example, uₙ = 2n − 1 gives every odd number instantly.
A recursive rule defines each term using the previous term. For example, t₁ = 1, tₙ = tₙ₋₁ + 3 for n ≥ 2. To use it, you must start from the first term and calculate step by step. The advantage is that it captures sequences which grow naturally from prior terms — like the Virahānka–Fibonacci sequence, where V₁ = 1, V₂ = 2, and Vₙ = Vₙ₋₁ + Vₙ₋₂ for n ≥ 3, giving 1, 2, 3, 5, 8, 13, 21, 34, …

Arithmetic Progressions (AP)

An AP is a sequence where consecutive terms differ by a constant amount called the common difference (d). It can increase (d > 0), decrease (d < 0), or stay flat (d = 0). The nth term is tₙ = a + (n − 1)d. When the pairs (n, tₙ) are plotted on a graph, they always fall on a straight line — a visual signature of any AP.

Sum of First n Natural Numbers

Using Āryabhaṭa’s method of writing the sum forwards and backwards, the formula Sₙ = n(n + 1)/2 is derived. This same formula gives the nth triangular number, since each triangular number is the sum of natural numbers up to that position.

Geometric Progressions (GP)

A GP is a sequence where each term is obtained by multiplying the previous term by a fixed common ratio (r). The nth term is tₙ = arⁿ⁻¹. Unlike an AP, a GP grows exponentially — when plotted, its terms do not lie on a straight line but form a curve. If r > 1 the terms grow rapidly; if 0 < r < 1 they shrink toward zero; if r is negative the terms alternate in sign.
The Sierpiński triangle is a striking real-world illustration: the number of black triangles at each stage follows a GP with r = 3 (growing fast), while the shaded area follows a GP with r = 3/4 (shrinking toward zero) — both operating simultaneously on the same figure.

Questions for Exam Point of View

These are the most important question types from Ganita Manjari Grade 9 Chapter 8. Students should be able to solve each type confidently before the exam.

Type 1: Finding the nth Term of an AP
These are direct one or two-mark questions and appear in almost every exam.
- Find the 20th term of the AP: 3, 8, 13, 18, …
  Here a = 3, d = 5. Using tₙ = a + (n − 1)d:
  t₂₀ = 3 + (20 − 1) × 5 = 3 + 95 = 98
- Find the nth term of the AP: 11, 8, 5, 2, …
  Here a = 11, d = −3. So tₙ = 11 + (n − 1)(−3) = 11 − 3n + 3 = 14 − 3n
Type 2: Which Term of an AP Equals a Given Value?
- Which term of the AP: 21, 18, 15, … is −81?
  Here a = 21, d = −3. Set tₙ = −81:
  21 + (n − 1)(−3) = −81
  (n − 1)(−3) = −102
  n − 1 = 34
  n = 35 — the 35th term is −81.
Type 3: Finding an AP from Two Conditions
- An AP has 50 terms. The 3rd term is 12 and the last term is 106. Find the 29th term.
  Two equations: a + 2d = 12 and a + 49d = 106.
  Subtracting: 47d = 94 → d = 2. Then a = 8.
  t₂₉ = 8 + (29 − 1) × 2 = 8 + 56 = 64
Type 4: Real-Life AP Problems
- Harish started work with 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?
  Here a = 5,00,000, d = 20,000, tₙ = 7,00,000.
  5,00,000 + (n − 1) × 20,000 = 7,00,000
  (n − 1) × 20,000 = 2,00,000
  n − 1 = 10
  n = 11 — after 11 years (i.e., at the start of the 11th year).
Type 5: Sum of Natural Numbers Formula
- How many 2-digit numbers are divisible by 3? Find their sum.
  The 2-digit multiples of 3 are 12, 15, 18, …, 99. This is an AP with a = 12, d = 3, last term = 99.
  Using 99 = 12 + (n − 1) × 3 → n = 30. There are 30 such numbers.
  Sum = (n/2)(first + last) = (30/2)(12 + 99) = 15 × 111 = 1665
Type 6: Finding the nth Term of a GP
- Find the 10th term of the GP: 5, 25, 125, …
  Here a = 5, r = 5. Using tₙ = arⁿ⁻¹:
  t₁₀ = 5 × 5⁹ = 5¹⁰ = 9,765,625
- Find the nth term of: 5, 25, 125, …
  tₙ = 5 × 5ⁿ⁻¹ = 5ⁿ
Type 7: Verifying a GP and Finding Common Ratio
- Is the sequence 2, 10, 50, 250, … a GP?
  Check ratios: 10/2 = 5, 50/10 = 5, 250/50 = 5. All ratios are equal, so yes, it is a GP with r = 5.
  nth term = 2 × 5ⁿ⁻¹
Type 8: Checking Membership in a Sequence
- Is 308 a term of the sequence with explicit rule sₙ = 5n − 2?
  Set 5n − 2 = 308 → 5n = 310 → n = 62. Since 62 is a natural number, yes, 308 is the 62nd term.
- Is 471 a term of the same sequence?
  Set 5n − 2 = 471 → 5n = 473 → n = 94.6. Since 94.6 is not a natural number, 471 is not a term of this sequence.
Type 9: Writing Explicit and Recursive Rules Together
- Write both the explicit rule and the recursive rule for the AP: 4, 9, 14, 19, …
  Here a = 4, d = 5.
  Explicit: tₙ = 4 + (n − 1) × 5 = 5n − 1
  Recursive: t₁ = 4, tₙ = tₙ₋₁ + 5 for n ≥ 2
Type 10: Sum Formula Application
- Find the sum: 25 + 26 + 27 + … + 58
  Using Sₙ = n(n + 1)/2:
  = S₅₈ − S₂₄ = (58 × 59)/2 − (24 × 25)/2 = 1711 − 300 = 1411.