Sequences are ordered lists of numbers where each number is called a term. Some sequences follow a rule, which helps students predict later terms without writing every earlier term.
Important Questions Class 9 Maths Chapter 8 help students practise explicit rules, recursive rules, arithmetic progressions, geometric progressions, triangular numbers, sum of natural numbers and pattern-based problems. This chapter is useful because many real-life situations grow by adding a fixed value or multiplying by a fixed value.
Class 9 Maths Chapter 8 begins with patterns students already know, such as natural numbers, odd numbers, triangular numbers and square numbers. It then explains how mathematicians use rules to predict what comes next in a sequence.
The chapter becomes important when students learn two powerful ideas. An arithmetic progression grows by adding the same number each time. A geometric progression grows by multiplying by the same number each time. These ideas appear in salaries, taxi fares, bouncing balls, bacteria growth, fractals and many exam questions.
Key Takeaways from Class 9 Maths Chapter 8
| Topic |
What Students Must Know |
| Chapter Name |
Predicting What Comes Next: Exploring Sequences and Progressions |
| Main Concept |
Ordered number patterns |
| Sequence |
Ordered list of terms |
| Explicit Rule |
Finds a term directly using its position |
| Recursive Rule |
Finds a term using previous term or terms |
| AP |
Sequence with common difference |
| GP |
Sequence with common ratio |
| Important Formula |
Sum of first (n) natural numbers = (\frac{n(n+1)}{2}) |
Important Questions Class 9 Maths Chapter 8 with Answers
These questions cover the basic ideas of the chapter. Students should learn the difference between term, position, explicit rule and recursive rule.
Important Questions Class 9 Maths Chapter 8: Sequence Basics
Q1. What is a sequence?
A sequence is an ordered list of numbers.
Each number in a sequence is called a term. For example, in (1, 4, 9, 16, 25, ...), the first term is 1, the second term is 4, and the fifth term is 25.
Q2. What is a finite sequence?
A finite sequence has a fixed number of terms.
For example:
6, 12, 24, 48, 96
This sequence has five terms, so it is finite.
Q3. What is an infinite sequence?
An infinite sequence continues without ending.
For example:
1, 2, 3, 4, 5, ...
This sequence of natural numbers continues indefinitely.
Q4. What is an explicit rule for a sequence?
An explicit rule gives the value of the nth term directly using (n).
For example:
u_n = 2n - 1
This gives the sequence of odd numbers:
1, 3, 5, 7, 9, ...
Q5. What is a recursive rule for a sequence?
A recursive rule defines a term using one or more previous terms.
For example:
t_1 = 1,  t_n = t_n-1 + 3
This gives the sequence:
1, 4, 7, 10, 13, ...
Class 9 Maths Chapter 8 Question Answer on Explicit Rules
Explicit rule class 9 questions are useful because students can find any term directly. They do not need to write all previous terms.
Explicit Rule Class 9 Questions
Q1. Find the first five terms of (t_n = 3n - 4).
Substitute (n = 1, 2, 3, 4, 5).
t_1 = 3(1) - 4 = -1
t_2 = 3(2) - 4 = 2
t_3 = 3(3) - 4 = 5
t_4 = 3(4) - 4 = 8
t_5 = 3(5) - 4 = 11
So, the first five terms are:
-1, 2, 5, 8, 11
Q2. Find the first five terms of (t_n = 2 - 5n).
t_1 = 2 - 5 = -3
t_2 = 2 - 10 = -8
t_3 = 2 - 15 = -13
t_4 = 2 - 20 = -18
t_5 = 2 - 25 = -23
So, the first five terms are:
-3, -8, -13, -18, -23
Q3. Find the 10th and 15th terms of (t_n = 5n - 3).
For the 10th term:
t_10 = 5(10) - 3
= 50 - 3 = 47
For the 15th term:
t_15 = 5(15) - 3
= 75 - 3 = 72
So, the 10th term is 47, and the 15th term is 72.
Q4. Determine whether 97 is a term of (t_n = 5n - 3).
Set:
5n - 3 = 97
5n = 100
n = 20
Since (n) is a natural number, 97 is the 20th term.
Q5. Determine whether 172 is a term of (t_n = 5n - 3).
Set:
5n - 3 = 172
5n = 175
n = 35
Since (n) is a natural number, 172 is the 35th term.
Recursive Rule Class 9 Important Questions
Recursive rule class 9 questions need earlier terms. Students should calculate each term step by step instead of jumping directly.
Q1. Find the first five terms of (t_1 = -5, t_n+1 = t_n + 3).
Given:
t_1 = -5
t_2 = -5 + 3 = -2
t_3 = -2 + 3 = 1
t_4 = 1 + 3 = 4
t_5 = 4 + 3 = 7
So, the first five terms are:
-5, -2, 1, 4, 7
Q2. Is 52 a term of (t_1 = -5, t_n+1 = t_n + 3)?
This sequence is:
-5, -2, 1, 4, 7, ...
Its nth term is:
t_n = -5 + (n - 1)3
Set:
-5 + 3(n - 1) = 52
3(n - 1) = 57
n - 1 = 19
n = 20
So, 52 is the 20th term.
Q3. Find the first four terms of (u_1 = 1, u_n = 2u_n-1 + 3).
u_1 = 1
u_2 = 2(1) + 3 = 5
u_3 = 2(5) + 3 = 13
u_4 = 2(13) + 3 = 29
So, the first four terms are:
1, 5, 13, 29
Q4. Find the first four terms of (s_1 = 3, s_n = s_n-1(s_n-1 - 1)).
s_1 = 3
s_2 = 3(3 - 1) = 6
s_3 = 6(6 - 1) = 30
s_4 = 30(30 - 1) = 870
So, the first four terms are:
3, 6, 30, 870
Q5. Find (T_4, T_5, T_6, T_7, T_8), if (T_1 = 1, T_2 = 2, T_3 = 4), and (T_n = T_n-1 + T_n-2 + T_n-3).
T_4 = T_3 + T_2 + T_1 = 4 + 2 + 1 = 7
T_5 = T_4 + T_3 + T_2 = 7 + 4 + 2 = 13
T_6 = T_5 + T_4 + T_3 = 13 + 7 + 4 = 24
T_7 = T_6 + T_5 + T_4 = 24 + 13 + 7 = 44
T_8 = T_7 + T_6 + T_5 = 44 + 24 + 13 = 81
Arithmetic Progression Class 9 Important Questions
An arithmetic progression, or AP, has a fixed difference between consecutive terms. This fixed difference is called the common difference.
Q1. What is an arithmetic progression?
An arithmetic progression is a sequence in which each term after the first is obtained by adding a fixed number to the previous term.
The fixed number is called the common difference.
Example:
3, 8, 13, 18, ...
Here, the common difference is 5.
Q2. What is the nth term of an AP?
The nth term of an AP is:
t_n = a + (n - 1)d
Here, (a) is the first term and (d) is the common difference.
Q3. Find the 10th and 26th terms of the AP (3, 8, 13, 18, ...).
Here:
a = 3,  d = 5
10th term:
t_10 = 3 + (10 - 1)5
= 3 + 45 = 48
26th term:
t_26 = 3 + (26 - 1)5
= 3 + 125 = 128
Q4. Which term of the AP (21, 18, 15, ...) is (-81)?
Here:
a = 21,  d = -3
Set:
t_n = -81
21 + (n - 1)(-3) = -81
21 - 3n + 3 = -81
24 - 3n = -81
-3n = -105
n = 35
So, (-81) is the 35th term.
Q5. Is 0 a term of the AP (21, 18, 15, ...)?
Set:
21 + (n - 1)(-3) = 0
24 - 3n = 0
3n = 24
n = 8
Since (n) is a natural number, 0 is the 8th term.
Sum of First n Natural Numbers Class 9 Questions
Sum of first n natural numbers class 9 questions are useful for triangular numbers and many word problems.
Class 9 Maths Chapter 8 Important Questions on Sums
Q1. What is the sum of the first (n) natural numbers?
The sum of the first (n) natural numbers is:
S_n = n(n + 1)/2
For example:
1 + 2 + 3 + ... + 10 = 10(11)/2 = 55
Q2. Find the sum of the first 25 natural numbers.
S_25 = 25(25 + 1)/2
= 25 × 26/2
= 25 × 13
= 325
Q3. Find (25 + 26 + 27 + ... + 58).
Use:
25 + 26 + ... + 58 = S_58 - S_24
S_58 = 58 × 59/2 = 1711
S_24 = 24 × 25/2 = 300
1711 - 300 = 1411
So, the sum is 1411.
Q4. A child arranges marbles in 25 rows. The first row has 1 marble, second has 2, and so on. Find the total marbles.
Total marbles:
1 + 2 + 3 + ... + 25
= 25(26)/2
= 325
So, the child uses 325 marbles.
Q5. Find the 10th, 17th and 80th triangular numbers.
The nth triangular number is:
T_n = n(n + 1)/2
10th triangular number:
T_10 = 10(11)/2 = 55
17th triangular number:
T_17 = 17(18)/2 = 153
80th triangular number:
T_80 = 80(81)/2 = 3240
Geometric Progression Class 9 Important Questions
A geometric progression, or GP, grows by multiplication. Students should check ratios, not differences.
Q1. What is a geometric progression?
A geometric progression is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed number.
The fixed number is called the common ratio.
Example:
3, 6, 12, 24, ...
Here, the common ratio is 2.
Q2. What is the nth term of a GP?
The nth term of a GP is:
t_n = ar^(n-1)
Here, (a) is the first term and (r) is the common ratio.
Q3. Find the 10th and nth terms of the GP (5, 25, 125, ...).
Here:
a = 5,  r = 5
nth term:
t_n = 5 × 5^(n-1)
= 5^n
10th term:
t_10 = 5^(10)
= 9765625
Q4. Which term of the GP (2, 6, 18, ...) is 4374?
Here:
a = 2,  r = 3
t_n = 2 × 3^(n-1)
Set:
2 × 3^(n-1) = 4374
3^(n-1) = 2187
2187 = 3^7
n - 1 = 7
n = 8
So, 4374 is the 8th term.
Q5. Find the 12th term of a GP with common ratio 2 and 8th term 192.
Given:
t_8 = 192,  r = 2
In a GP:
t_12 = t_8 × r^(12-8)
= 192 × 2^4
= 192 × 16
= 3072
So, the 12th term is 3072.
Fractal Sequences in Predicting What Comes Next Class 9
Fractal sequences make this chapter more visual and different from a regular AP-GP chapter. They show how patterns can repeat at smaller and smaller scales.
Q1. How does the Sierpiński triangle form a sequence?
In the Sierpiński triangle, each stage creates smaller copies of the earlier shape.
The number of black triangles follows:
1, 3, 9, 27, 81, ...
This is a GP with common ratio 3.
Q2. What is the number of black triangles at Stage (n) in the Sierpiński triangle?
The rule is:
t_n = 3^n
So, at Stage 5:
t_5 = 3^5 = 243
Q3. How does the black area change in the Sierpiński triangle?
At each stage, the black area becomes (3/4) of the previous stage.
So, if the starting area is 1 square unit, the black area at Stage (n) is:
(3/4)^n
Q4. What happens in the Sierpiński square carpet?
In the Sierpiński square carpet, a square is divided into 9 equal smaller squares and the centre square is removed.
This process repeats for each remaining square. It creates a sequence based on repeated multiplication.
Q5. Why are fractal sequences useful in this chapter?
Fractal sequences show that patterns are not limited to simple lists of numbers.
They connect geometry with progressions. Students see how repeated rules can create complex shapes.
Class 9 Maths Chapter 8 MCQ with Answers
Class 9 Maths Chapter 8 MCQs test sequence rules, AP, GP and formula recall. These are useful for quick revision.
Q1. A sequence is:
(a) An unordered list
(b) An ordered list of numbers
(c) A single number
(d) A shape
Answer: (b) An ordered list of numbers
A sequence has terms arranged in a specific order.
Q2. The common difference of (3, 8, 13, 18, ...) is:
(a) 3
(b) 4
(c) 5
(d) 8
Answer: (c) 5
Each term increases by 5.
Q3. The nth term of an AP is:
(a) (ar^(n-1))
(b) (a + (n - 1)d)
(c) (n(n+1)/2)
(d) (a^2 + b^2)
Answer: (b) (a + (n - 1)d)
This formula gives the nth term of an AP.
Q4. The common ratio of (3, 6, 12, 24, ...) is:
(a) 2
(b) 3
(c) 4
(d) 6
Answer: (a) 2
Each term is multiplied by 2.
Q5. The nth term of a GP is:
(a) (a + (n - 1)d)
(b) (ar^(n-1))
(c) (n^2)
(d) (2n - 1)
Answer: (b) (ar^(n-1))
This formula gives the nth term of a GP.
Q6. The sequence (1, 3, 9, 27, ...) is:
(a) AP
(b) GP
(c) Neither AP nor GP
(d) A finite sequence only
Answer: (b) GP
Each term is multiplied by 3.
Class 9 Maths Chapter 8 Extra Questions with Answers
Class 9 maths chapter 8 extra questions help students practise AP, GP, recursive rules and real-life patterns.
Q1. Find the nth term of the AP (11, 8, 5, 2, ...).
Here:
a = 11,  d = -3
t_n = a + (n - 1)d
t_n = 11 + (n - 1)(-3)
= 11 - 3n + 3
= 14 - 3n
So, the nth term is:
t_n = 14 - 3n
Q2. Write the recursive rule for the AP (11, 8, 5, 2, ...).
The first term is 11.
Each term decreases by 3.
So:
t_1 = 11,  t_n = t_n-1 - 3, n ≥ 2
Q3. How many two-digit numbers are divisible by 3?
The first two-digit multiple of 3 is 12.
The last two-digit multiple of 3 is 99.
This forms an AP:
12, 15, 18, ..., 99
Here:
a = 12,  d = 3,  t_n = 99
12 + (n - 1)3 = 99
(n - 1)3 = 87
n - 1 = 29
n = 30
So, there are 30 two-digit numbers divisible by 3.
Q4. Find the sum of all two-digit numbers divisible by 3.
The numbers are:
12, 15, 18, ..., 99
There are 30 terms.
Sum:
S = n/2(a + l)
= 30/2(12 + 99)
= 15 × 111
= 1665
So, the sum is 1665.
Q5. Harish starts with an annual salary of ₹5,00,000 and gets ₹20,000 increment each year. After how many years will his income reach ₹7,00,000?
This forms an AP:
500000, 520000, 540000, ...
Here:
a = 500000,  d = 20000
Set:
a + (n - 1)d = 700000
500000 + (n - 1)20000 = 700000
(n - 1)20000 = 200000
n - 1 = 10
n = 11
So, his salary reaches ₹7,00,000 in the 11th year.
Application-Based Questions from Predicting What Comes Next Class 9
Application questions show why sequences matter. Read the situation carefully to decide whether it uses AP or GP.
Q1. A taxi company charges ₹200 fixed fee and ₹40 per km. Find the fare after 10 km.
Fare after (n) km:
200 + 40n
For 10 km:
200 + 40(10)
= 200 + 400
= 600
So, the fare is ₹600.
Q2. A growing square pattern has (1, 5, 9, 13, ...) squares. Find the 20th term.
This is an AP.
a = 1,  d = 4
t_n = 1 + (n - 1)4
t_n = 4n - 3
For (n = 20):
t_20 = 4(20) - 3
= 80 - 3 = 77
So, Stage 20 has 77 squares.
Q3. A bacteria culture starts with 30 bacteria and doubles every hour. Find bacteria at the end of 2nd, 4th and nth hour.
The number of bacteria forms a GP.
Original number = 30.
After 1 hour:
30 × 2
After 2 hours:
30 × 2^2 = 120
After 4 hours:
30 × 2^4 = 480
After (n) hours:
30 × 2^n
Q4. A ball is dropped from 80 m and bounces to 60% of the previous height. Find the height after the 5th bounce.
Each bounce height forms a GP.
First bounce:
80 × 0.6
Height after 5th bounce:
80 × (0.6)^5
= 80 × 0.07776
= 6.2208
So, the height after the 5th bounce is 6.2208 m.
Q5. In the Sierpiński triangle, how many black triangles are there at Stage 5?
The number of black triangles is:
3^n
At Stage 5:
3^5 = 243
So, Stage 5 has 243 black triangles.
Long Answer Questions from Class 9 Maths Chapter 8
Long answers should show the rule, substitution and final result clearly.
Q1. Explain the difference between explicit and recursive rules.
An explicit rule gives the value of a term directly using its position number.
For example:
t_n = 3n - 2
This gives the nth term without finding earlier terms.
A recursive rule gives a term using one or more previous terms.
For example:
t_1 = 1,  t_n = t_n-1 + 3
Here, we need the previous term to find the next term.
Explicit rules save time for far-away terms. Recursive rules show how a sequence grows step by step.
Q2. Explain how to identify an arithmetic progression.
A sequence is an arithmetic progression if the difference between consecutive terms is constant.
For example:
2, 5, 8, 11, ...
The differences are:
5 - 2 = 3
8 - 5 = 3
11 - 8 = 3
Since the difference stays 3, this sequence is an AP.
The nth term of an AP is:
t_n = a + (n - 1)d
Q3. Explain how to identify a geometric progression.
A sequence is a geometric progression if the ratio of consecutive terms is constant.
For example:
3, 6, 12, 24, ...
The ratios are:
6/3 = 2
12/6 = 2
24/12 = 2
Since the ratio stays 2, this sequence is a GP.
The nth term of a GP is:
t_n = ar^(n-1)
Q4. Derive the formula for the sum of the first (n) natural numbers.
Let:
S = 1 + 2 + 3 + ... + n
Write the same sum in reverse order:
S = n + (n - 1) + (n - 2) + ... + 1
Add both:
2S = (n + 1) + (n + 1) + ... + (n + 1)
There are (n) such pairs.
2S = n(n + 1)
S = n(n + 1)/2
So:
1 + 2 + 3 + ... + n = n(n + 1)/2
Q5. Explain how the Sierpiński triangle gives a geometric progression.
In the Sierpiński triangle, Stage 0 has 1 black triangle.
At each new stage, every black triangle gets replaced by 3 smaller black triangles.
So, the number of black triangles is:
1, 3, 9, 27, 81, ...
This is a GP because each term is obtained by multiplying the previous term by 3.
The rule for Stage (n) is:
t_n = 3^n
If the starting area is 1 square unit, the black area at each stage is multiplied by (3/4).
So, the area at Stage (n) is:
(3/4)^n