πŸ”„ CBSE Class 9 Mathematics Β· 2026–27 New Syllabus Β· Chapter 6 NEW

Sequences &
Progressions

Complete visual guide to sequences, Arithmetic Progressions (AP), Geometric Progressions (GP), nth term, sum formulas, graphs, fractals, and the Tower of Hanoi. Entirely new to Class 9 in 2026-27. Chapter 6 Β· 7 Marks.

πŸ“Œ Chapter 6 Β· 7 Marks
πŸ†• NEW in Class 9
Sequences
Arithmetic Progression (AP)
AP Sum Formula
Geometric Progression (GP)
Fractals
50 Practice Qs
Section 6.1
What is a Sequence?
A sequence is a list of numbers in a definite order β€” each number follows a specific rule. Understanding sequences is the foundation of this entire chapter.
πŸ“Œ Definition β€” Sequence
A sequence is an ordered list of numbers (called terms) where each term follows a rule. The terms are denoted a₁, aβ‚‚, a₃, … or T₁, Tβ‚‚, T₃ …

Finite sequence: has a fixed number of terms β€” e.g., 2, 4, 6, 8, 10
Infinite sequence: continues indefinitely β€” e.g., 1, 4, 9, 16, 25, …

Types of Rules for Sequences

TypeHow to find next termExample
RecursiveEach term defined using previous term(s)aβ‚™ = aₙ₋₁ + 3
Explicitnth term defined directly using naβ‚™ = 2n + 1
General TermFormula for any term β€” most useful!Tβ‚™ = a + (nβˆ’1)d

πŸ“ Arithmetic Sequence

Each term is obtained by adding a fixed number to the previous term.

2
+3
5
+3
8
+3
11

Common difference d = 3 (constant addition)

πŸ“ Geometric Sequence

Each term is obtained by multiplying a fixed number to the previous term.

2
Γ—3
6
Γ—3
18
Γ—3
54

Common ratio r = 3 (constant multiplication)

Section 6.2
Arithmetic Progressions (AP)
An AP is a sequence where the difference between consecutive terms is constant. This "common difference" can be positive, negative, or zero.
πŸ“Œ Definition β€” Arithmetic Progression
A sequence a₁, aβ‚‚, a₃, … is an Arithmetic Progression (AP) if:
aβ‚‚ βˆ’ a₁ = a₃ βˆ’ aβ‚‚ = aβ‚„ βˆ’ a₃ = … = constant = d (common difference)

Standard form: a, a+d, a+2d, a+3d, …
where a = first term and d = common difference
πŸ“ nth Term Formula β€” Arithmetic Progression
Tβ‚™ = a + (n βˆ’ 1)d
where: a = first term  |  d = common difference  |  n = term number
Note: T₁ = a, Tβ‚‚ = a+d, T₃ = a+2d, … Last term: l = a + (nβˆ’1)d
n Tβ‚™ 123 45 T₁=5Tβ‚‚=20T₃=35 Tβ‚„=50Tβ‚…=65 d=15 AP: a=5, d=15 β†’ straight line graph
Graph of an AP: a=5, d=15. The points lie on a straight line β€” characteristic of arithmetic progressions.
Example 1
For the AP 7, 13, 19, 25, … find (a) first term (b) common difference (c) 15th term. [3 Marks]
a
First term a = 7
b
Common difference d = Tβ‚‚ βˆ’ T₁ = 13 βˆ’ 7 = 6
Check: 19βˆ’13=6 βœ“, 25βˆ’19=6 βœ“
c
15th term: Tβ‚™ = a + (nβˆ’1)d
T₁₅ = 7 + (15βˆ’1)Γ—6 = 7 + 84 = 91
a = 7, d = 6, T₁₅ = 91
Example 2
Which term of the AP 5, 11, 17, 23, … is 65? [2 Marks]
1
a = 5, d = 6, Tβ‚™ = 65
2
Use: Tβ‚™ = a + (nβˆ’1)d
65 = 5 + (nβˆ’1)Γ—6
60 = (nβˆ’1)Γ—6
nβˆ’1 = 10 β†’ n = 11
65 is the 11th term
Example 3
The 5th and 10th terms of an AP are 23 and 43. Find a, d, and the 20th term. [3 Marks]
1
Tβ‚… = a + 4d = 23   ...(i)
T₁₀ = a + 9d = 43   ...(ii)
2
Subtract (i) from (ii): 5d = 20 β†’ d = 4
3
From (i): a + 16 = 23 β†’ a = 7
4
Tβ‚‚β‚€ = 7 + 19Γ—4 = 7 + 76 = 83
a = 7, d = 4, Tβ‚‚β‚€ = 83
Section 6.3
Sum of an Arithmetic Progression
The sum of the first n terms of an AP is denoted Sβ‚™. Gauss famously derived this formula as a child β€” summing 1 to 100 instantly.
πŸ“ Sum of First n Terms of an AP
Sβ‚™ = n/2 Γ— [2a + (nβˆ’1)d]
Alternate form when last term l is known: Sβ‚™ = n/2 Γ— (a + l)
where l = last term = a + (nβˆ’1)d
Also useful: Tβ‚™ = Sβ‚™ βˆ’ Sₙ₋₁ (nth term from sum)
Gauss's Trick: Sum 1+2+3+…+100 = ? 1+2 +3 + … +98 +99 +100 100+99 +98 + … +3 +2 +1 Each pair = 101 100 pairs β†’ 2S = 100Γ—101 S = 100Γ—101/2 = 5050
Gauss's pairing trick: adding AP forwards + backwards gives n equal pairs, each summing to (a+l)
Example 4
Find the sum of the first 20 terms of AP: 3, 7, 11, 15, … [2 Marks]
1
a = 3, d = 4, n = 20
2
Sβ‚‚β‚€ = n/2 Γ— [2a + (nβˆ’1)d]
= 20/2 Γ— [2(3) + 19Γ—4]
= 10 Γ— [6 + 76]
= 10 Γ— 82 = 820
Sβ‚‚β‚€ = 820
Example 5
Find the sum of all multiples of 7 between 1 and 100. [3 Marks]
1
Multiples of 7 between 1 and 100: 7, 14, 21, … , 98
This is an AP with a = 7, d = 7, l = 98
2
Find n: l = a+(nβˆ’1)d β†’ 98 = 7+(nβˆ’1)Γ—7 β†’ 91 = (nβˆ’1)Γ—7 β†’ n = 14
3
Sβ‚™ = n/2 Γ— (a+l) = 14/2 Γ— (7+98) = 7 Γ— 105 = 735
Sum of multiples of 7 between 1 and 100 = 735
Section 6.4 β€” NEW in Class 9 (from Cl.11)
Geometric Progressions (GP)
A GP is a sequence where each term is obtained by multiplying the previous term by a fixed number called the common ratio r.
πŸ“Œ Definition β€” Geometric Progression
A sequence a₁, aβ‚‚, a₃, … is a Geometric Progression (GP) if:
aβ‚‚/a₁ = a₃/aβ‚‚ = aβ‚„/a₃ = … = constant = r (common ratio)

Standard form: a, ar, arΒ², arΒ³, …
where a = first term and r = common ratio (r β‰  0)
πŸ“ nth Term Formula β€” Geometric Progression
Tβ‚™ = a Β· rⁿ⁻¹
T₁ = a, Tβ‚‚ = ar, T₃ = arΒ², Tβ‚„ = arΒ³, …
If r > 1: sequence grows (exponential growth)
If 0 < r < 1: sequence shrinks (converges to 0)
If r = 1: all terms equal a (constant sequence)
If r < 0: alternating positive/negative

πŸ“ˆ GP with r > 1 (Growing)

2
Γ—2
4
Γ—2
8
Γ—2
16

a=2, r=2. Doubles each time. Exponential growth.

πŸ“‰ GP with 0 < r < 1 (Shrinking)

64
Γ·2
32
Γ·2
16
Γ·2
8

a=64, r=Β½. Halves each time β†’ approaches 0.

Example 6
For the GP 3, 6, 12, 24, … find (a) r (b) Tβ‚ˆ (c) Is 768 a term? [3 Marks]
a
r = Tβ‚‚/T₁ = 6/3 = 2
b
Tβ‚ˆ = aΒ·r⁷ = 3Γ—2⁷ = 3Γ—128 = 384
c
If 768 = 3Γ—2ⁿ⁻¹ β†’ 2ⁿ⁻¹ = 256 = 2⁸ β†’ nβˆ’1=8 β†’ n=9
768 = T₉ βœ“
r = 2, Tβ‚ˆ = 384, 768 = T₉ (yes, it is a term)
⚠️
Common Mistake: Always check whether r is calculated as Tβ‚‚/T₁ (not T₁/Tβ‚‚). Also, r cannot be 0 in a GP. If r = 1, every term equals a β€” a valid but trivial GP.
Section 6.5
Applications: Fractals & Tower of Hanoi
The 2026-27 syllabus includes beautiful real-world and puzzle applications of sequences. These are tested in case-based questions.

🌿 Fractals β€” Sequences in Nature

A fractal is a pattern that repeats itself at every scale. The SierpiΕ„ski Triangle is a classic fractal built using an AP/GP pattern.

Step 0: 1 triangle Step 1: 3 triangles Step 2: 9 triangles Pattern: 1, 3, 9, 27, … β†’ GP with r = 3

At each step, each triangle splits into 3 smaller ones. The number of triangles forms a GP: 1, 3, 9, 27, … with first term 1 and r = 3. After n steps: 3ⁿ triangles.

πŸ—Ό Tower of Hanoi β€” Sequences in Puzzles

The Tower of Hanoi puzzle with n discs requires a minimum of 2ⁿ βˆ’ 1 moves. This forms a sequence of minimum moves:

n=112ΒΉβˆ’1 n=232Β²βˆ’1 n=372Β³βˆ’1 n=4152β΄βˆ’1 n=5312β΅βˆ’1 Sequence: 1, 3, 7, 15, 31, … β†’ GP-like: each term = 2Γ—previous + 1

The moves sequence 1, 3, 7, 15, 31 is actually 2ⁿ βˆ’ 1. It's related to a GP: the differences form a GP (1, 2, 4, 8, …). The legend says a monk moving 64 golden discs would take 2βΆβ΄βˆ’1 β‰ˆ 18 quintillion moves!

NCERT Style Β· CBSE Pattern
More Worked Examples
These cover important exam patterns combining AP and GP concepts.
Example 7
The sum of first 10 terms of an AP is 175 and the sum of first 5 terms is 50. Find a and d. [3 Marks]
1
S₁₀ = 10/2[2a+9d] = 5[2a+9d] = 175 β†’ 2a+9d = 35 …(i)
2
Sβ‚… = 5/2[2a+4d] = 5(a+2d)/... Actually: Sβ‚… = 5/2[2a+4d] = 5[a+2d] = 50 β†’ a+2d = 10 …(ii)
3
From (ii): a = 10βˆ’2d. Sub in (i): 2(10βˆ’2d)+9d=35 β†’ 20+5d=35 β†’ d=3, a=10βˆ’6=4
a = 4, d = 3
Example 8
Three numbers are in GP. Their product is 216 and sum is 19. Find the numbers. [3 Marks]
1
Let the three numbers be a/r, a, ar (this makes product and sum easier)
2
Product: (a/r)Β·aΒ·(ar) = aΒ³ = 216 β†’ a = 6
3
Sum: 6/r + 6 + 6r = 19 β†’ 6/r + 6r = 13 β†’ 6rΒ² βˆ’ 13r + 6 = 0
β†’ (2rβˆ’3)(3rβˆ’2)=0 β†’ r = 3/2 or r = 2/3
4
If r=3/2: numbers = 4, 6, 9. If r=2/3: numbers = 9, 6, 4 (same set)
The three numbers are 4, 6, 9
NCERT Exercise
Exercise 6.1 & 6.2 β€” Solved
Standard CBSE exercise questions with complete step-by-step working.
Q1 Β· 2 Marks

For the AP: 2, 7, 12, 17, … find the 31st term.

1
a=2, d=7βˆ’2=5, n=31
2
T₃₁ = a+(nβˆ’1)d = 2+30Γ—5 = 2+150 = 152
T₃₁ = 152
Q2 Β· 3 Marks

Find the sum: 1 + 3 + 5 + 7 + … + 99 (sum of first 50 odd numbers)

1
AP with a=1, d=2, l=99. Find n: 99=1+(nβˆ’1)Γ—2 β†’ n=50
2
Sβ‚…β‚€ = 50/2Γ—(1+99) = 25Γ—100 = 2500
Sum = 2500 (= 50Β²: sum of first n odd numbers = nΒ²)
Q3 Β· 2 Marks

For the GP 5, 25, 125, … find the 6th term.

1
a=5, r=25/5=5
2
T₆ = 5Γ—5⁡ = 5Γ—3125 = 15625
T₆ = 15625
Q4 Β· 3 Marks

The 4th term of a GP is 8 and the 8th term is 128. Find the first term and common ratio.

1
Tβ‚„ = arΒ³ = 8 …(i)   Tβ‚ˆ = ar⁷ = 128 …(ii)
2
Divide (ii)Γ·(i): r⁴ = 128/8 = 16 β†’ r = 2
3
From (i): aΓ—8 = 8 β†’ a = 1
a = 1, r = 2. GP: 1, 2, 4, 8, 16, 32, 64, 128 …
Smart Study
10 Study Tips for Sequences & Progressions
Targeted tips for the most common CBSE exam mistakes in this chapter.
1

Always find d = Tβ‚‚ βˆ’ T₁, not T₁ βˆ’ Tβ‚‚

Common difference d is always the LATER term minus the EARLIER term. d = Tβ‚‚βˆ’T₁. If you reverse it, you'll get the wrong sign.

2

Sβ‚™ βˆ’ Sₙ₋₁ = Tβ‚™

The nth term equals the sum of n terms minus the sum of (nβˆ’1) terms. This is tested in 2-mark questions β€” don't forget it.

3

For GP: r = Tβ‚‚/T₁ (not T₁/Tβ‚‚)

Common ratio r = any term Γ· previous term. Always check r by trying T₃/Tβ‚‚ = Tβ‚‚/T₁.

4

Three AP terms trick: aβˆ’d, a, a+d

If three numbers are in AP, let them be (aβˆ’d), a, (a+d). Their sum = 3a and product is easy. Saves algebra.

5

Three GP terms trick: a/r, a, ar

If three numbers are in GP, let them be a/r, a, ar. Their product = aΒ³. Saves algebra enormously (as in Example 8).

6

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

This is Sβ‚™ for AP: a=1, d=1. Formula: n(n+1)/2. Use directly when asked to sum 1+2+3+…+n.

7

Sum of first n odd numbers = nΒ²

1+3+5+…+(2nβˆ’1) = nΒ². Pure AP: a=1, d=2. Verified: 1+3+5=9=3Β². Very frequent in CBSE.

8

AP graph is a straight line; GP graph is exponential

Plot AP terms β†’ straight line (constant slope = d). Plot GP terms β†’ exponential curve. CBSE asks to identify the type from graph.

9

Tower of Hanoi: minimum moves = 2ⁿ βˆ’ 1

For n discs: 1 disc β†’ 1 move, 2 discs β†’ 3 moves, 3 discs β†’ 7 moves. General: 2ⁿ βˆ’ 1. Comes up in case-based questions.

10

Alternate sum formula: Sβ‚™ = n/2(a + l)

When you know first and last term, use Sβ‚™ = n/2(a+l). When you know a and d but not l, use Sβ‚™ = n/2[2a+(nβˆ’1)d].

Quick Reference

Chapter 6 β€” Complete Formula Sheet

ConceptFormulaNotes
AP β€” nth termTβ‚™ = a + (nβˆ’1)da=first term, d=common diff
AP β€” last terml = a + (nβˆ’1)dSame formula with l instead of Tβ‚™
AP β€” Sum (using d)Sβ‚™ = n/2[2a+(nβˆ’1)d]Use when d is known
AP β€” Sum (using l)Sβ‚™ = n/2(a+l)Use when last term l is known
nth term from sumTβ‚™ = Sβ‚™ βˆ’ Sₙ₋₁For n β‰₯ 2; T₁ = S₁
GP β€” nth termTβ‚™ = aΒ·rⁿ⁻¹a=first term, r=common ratio
Sum 1 to nn(n+1)/2AP: a=1, d=1
Sum first n oddsnΒ²AP: a=1, d=2
SierpiΕ„ski triangles3ⁿ at step nGP with r=3
Tower of Hanoi2ⁿ βˆ’ 1 for n discsRelated to GP
CBSE Pattern Practice
50 Practice Questions
MCQ Β· 1 Mark Β· 2 Marks Β· 3 Marks Β· 5 Marks Β· Case-Based β€” all with full solutions.