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

Class 9 Ganita Manjari Chapter 8 Quick Links:

• Exercise Set 8.1
• Exercise Set 8.2
• Exercise Set 8.3
• End-of-Chapter Exercises

To study with mobile or tab, download Class 9 Maths Ganita Manjari Offline App here.

The chapter 8 of Ganita Manjari class 9th also connects these ideas to real-life contexts — taxi fares, salary increments, bouncing balls, bacterial growth — and to beautiful mathematical objects like the Sierpiński triangle. For exam preparation, three things matter most: knowing the formulas cold, understanding when to apply each one and practising the variety of question types that appear in school and board-level assessments.

Class 9 Maths Ganita Manjari Chapter 8 Exercise Set 8.1 Solutions

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.

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.

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.

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

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.

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

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.

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.

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.

(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.

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.

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.

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

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.

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.

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.

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.

(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.

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.

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

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.

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.

(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.

(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.

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, …

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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:

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

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.

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, …

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.

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.

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.

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.

1. 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
2. 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.
3. 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
4. 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).
5. 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
6. 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
7. 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
8. 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.
9. 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
10. 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.