Exploring Sequences and Progressions Class 9 Notes and Solutions

Predicting What Comes Next: Exploring Sequences and Progressions

Patterns are everywhere — in nature, art, music, and even in the numbers we use every day. In mathematics, a sequence is a special kind of pattern made by numbers or objects arranged in a specific order. By understanding sequences, we can predict future values, solve real-life problems, and explore how numbers grow, shrink, or repeat.

Table of Contents

8.1 Introduction to Sequences

A sequence is simply an ordered list of numbers where each number is called a term. Sequences can be finite (having a limited number of terms) or infinite (continuing forever).

Imp Point

The three dots (…) at the end of a sequence mean the pattern continues indefinitely.

Here are some famous sequences you already know:

Sequence Name	Terms	Pattern Rule
Natural Numbers	1, 2, 3, 4, 5, 6, …	Add 1 to the previous term
Odd Numbers	1, 3, 5, 7, 9, 11, …	Add 2 to the previous term
Triangular Numbers	1, 3, 6, 10, 15, 21, …	Sum of natural numbers up to that position
Square Numbers	1, 4, 9, 16, 25, 36, …	Position number squared

Triangular Numbers Visual

Triangular numbers get their name because they can be arranged as triangles made of dots. Each term adds a new bottom row to the triangle.

The differences between consecutive triangular numbers are 2, 3, 4, 5, 6… which follows a clear pattern. Also, each triangular number equals the sum of natural numbers up to that point:

1st triangular number = 1
2nd triangular number = 1 + 2 = 3
3rd triangular number = 1 + 2 + 3 = 6
5th triangular number = 1 + 2 + 3 + 4 + 5 = 15

Square Numbers and Odd Numbers Connection

Square numbers have a beautiful relationship with odd numbers. Each square number is the sum of consecutive odd numbers starting from 1:

1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7

Imp Notation

We use t₁ for the first term, t₂ for the second term, and tₙ for the nth term of a sequence. The subscript always tells us the position of the term.

8.2 Explicit Rules for Sequences

An explicit formula (or explicit rule) uses the position number n to directly calculate the value of any term without knowing the previous terms.

For the sequence of odd numbers:

tₙ = 2n − 1

Substitute n = 1, 2, 3… to get 1, 3, 5, 7…

The power of an explicit formula is that you can find the 100th term or even the 1000th term instantly without listing all previous terms.

Example: Find the 53rd odd number.

t₅₃ = 2(53) − 1 = 106 − 1 = 105

You can also use explicit formulas to check if a number belongs to a sequence. For example, is 137 an odd number in this sequence?

Set tₙ = 137

2n − 1 = 137

2n = 138 → n = 69

Since n = 69 is a whole number, 137 is the 69th odd number.

Imp Rule

If solving for n gives a whole number (natural number), the value IS a term of the sequence. If n is a decimal or fraction, it is NOT a term.

8.3 Recursive Rules for Sequences

A recursive formula defines each term based on the previous term(s). You need to know earlier terms to find the next one.

Consider the sequence: 1, 4, 7, 10, 13, … Each term is 3 more than the previous term. The recursive rule is:

t₁ = 1

tₙ = tₙ₋₁ + 3 (for n ≥ 2)

Virahanka-Fibonacci Sequence

One of the most famous recursive sequences was discovered by the Indian mathematician Virahanka in the 7th century CE while studying poetry and meter. It was later studied by Hemachandra and the Italian mathematician Fibonacci.

V₁ = 1, V₂ = 1

Vₙ = Vₙ₋₁ + Vₙ₋₂ (for n ≥ 3)

This gives the sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, …

Each term is the sum of the two previous terms. This sequence appears everywhere in nature — from flower petals to spiral galaxies!

8.4 Arithmetic Progressions (AP)

An Arithmetic Progression (AP) is a sequence where the difference between any two consecutive terms is always the same. This constant difference is called the common difference (d).

General Form of an AP:

a, (a + d), (a + 2d), (a + 3d), …, [a + (n − 1)d]

Explicit Formula for the nth Term:

tₙ = a + (n − 1)d

Where a = first term, d = common difference, n = term position

Imp Property

When you plot the terms of an AP as points (position, value), they always lie on a straight line. This shows that an AP grows at a constant rate.

8.5 Sum of Natural Numbers

How can we find the sum 1 + 2 + 3 + … + 100 without adding each number? The great mathematician Carl Friedrich Gauss found a clever trick as a schoolboy, and Indian mathematician Aryabhata also described this method.

Write the sum forwards and backwards, then add them:

S = 1 + 2 + 3 + … + (n−1) + n

S = n + (n−1) + (n−2) + … + 2 + 1

2S = (n+1) + (n+1) + … [n times]

2S = n(n + 1)

S = n(n + 1)2

Imp Formula

The nth triangular number is exactly the sum of the first n natural numbers:

Tₙ = n(n + 1)2

This formula also helps find the sum of any consecutive number range. For example, to find 25 + 26 + … + 58:

Sum = (Sum of 1 to 58) − (Sum of 1 to 24)

Sum = 58 × 592 − 24 × 252 = 1711 − 300 = 1411

8.6 Geometric Progressions (GP)

A Geometric Progression (GP) is a sequence where each term is found by multiplying the previous term by a fixed number called the common ratio (r).

General Form of a GP:

a, ar, ar², ar³, …, arⁿ⁻¹

Explicit Formula for the nth Term:

tₙ = a × rⁿ⁻¹

Where a = first term, r = common ratio, n = term position

Imp Difference

In an AP, we add the common difference. In a GP, we multiply by the common ratio. AP grows linearly; GP grows exponentially.

Sierpinski Triangle and Fractals

A fractal is a pattern that repeats itself at different scales. The Sierpinski triangle is a famous fractal created by repeatedly removing the middle triangle.

The number of black triangles follows a GP: 1, 3, 9, 27, 81, … with common ratio r = 3.

The area of the black region also follows a GP: 1, 3/4, 9/16, 27/64, … with common ratio r = 3/4.

Stage (n)	Black Triangles	Shaded Area
0	1 = 3⁰	1
1	3 = 3¹	3/4
2	9 = 3²	9/16 = (3/4)²
3	27 = 3³	27/64 = (3/4)³
n	3ⁿ	(3/4)ⁿ

Imp Observation

As stages increase, the number of triangles grows very large, but the total shaded area becomes smaller and smaller, approaching zero. This is the magic of fractals — infinite detail in finite space!

Exercise Set 8.1 — Solutions

Question 1: First five terms

(i) tₙ = 3n − 4

Solution:

t₁ = −1; t₂ = 2; t₃ = 5; t₄ = 8; t₅ = 11

Sequence: −1, 2, 5, 8, 11

(ii) tₙ = 2 − 5n

Solution:

t₁ = −3; t₂ = −8; t₃ = −13; t₄ = −18; t₅ = −23

Sequence: −3, −8, −13, −18, −23

(iii) tₙ = n² − 2n + 3

Solution:

t₁ = 2; t₂ = 3; t₃ = 6; t₄ = 11; t₅ = 18

Sequence: 2, 3, 6, 11, 18

Question 2: 10th and 15th terms of tₙ = 5n − 3

Solution:

t₁₀ = 5(10) − 3 = 47

t₁₅ = 5(15) − 3 = 72

Question 3: Is 97 or 172 a term of tₙ = 5n − 3?

Solution:

For 97: 5n = 100 → n = 20. 97 is the 20th term.

For 172: 5n = 175 → n = 35. 172 is the 35th term.

Question 4: Which term is 607 in tₙ = 5n − 3?

Solution:

5n − 3 = 607 → 5n = 610 → n = 122. 607 is the 122nd term.

Question 5: Recursive sequence t₁ = −5, tₙ₊₁ = tₙ + 3

Solution:

t₁ = −5; t₂ = −2; t₃ = 1; t₄ = 4; t₅ = 7

For 52: AP with a = −5, d = 3 → −5 + (n−1)3 = 52 → n = 20. 52 is the 20th term.

Question 6: Tribonacci sequence — T₁=1, T₂=2, T₃=4, Tₙ = Tₙ₋₁+Tₙ₋₂+Tₙ₋₃

Solution:

T₄ = 7; T₅ = 13; T₆ = 24; T₇ = 44; T₈ = 81

Exercise Set 8.2 — Solutions

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

Solution:

a = 3, d = 5. tₙ = 5n − 2

t₁₀ = 48; t₂₆ = 128

Question 2: Which term of 21, 18, 15, … is −81? Is 0 a term?

Solution:

a = 21, d = −3. tₙ = 24 − 3n

−81 is the 35th term. 0 is the 8th term.

Question 3: nth term and recursive rule for AP 11, 8, 5, 2, …

Solution:

a = 11, d = −3

Explicit: tₙ = 14 − 3n

Recursive: t₁ = 11, tₙ = tₙ₋₁ − 3 (for n ≥ 2)

Question 4: 50-term AP, 3rd term = 12, last term = 106. Find 29th term.

Solution:

a + 2d = 12; a + 49d = 106 → 47d = 94 → d = 2; a = 8

t₂₉ = 8 + 28×2 = 64

Question 5: Two-digit numbers divisible by 3 — count and sum

Solution:

AP: 12, 15, …, 99 (a=12, d=3). n = 30.

Sum = 15 × 111 = 1665

Question 6: Salary from ₹5,00,000 with ₹20,000 yearly increment — when does it reach ₹7,00,000?

Solution:

500000 + (n−1)×20000 = 700000 → n = 11. Reached in the 11th year.

Question 7: Marbles in rows 1+2+…+25

Solution:

S = 25×26÷2 = 325 marbles

Exercise Set 8.3 — Solutions

Question 1: 12th term of GP with r = 2, 8th term = 192

Solution:

a×2⁷ = 192 → a = 3/2. t₁₂ = (3/2)×2¹¹ = 3072

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

Solution:

a=5, r=5. tₙ = 5ⁿ. t₁₀ = 5¹⁰ = 9,765,625

Question 3: t₁=2, tₙ₊₁=3tₙ−2. Which term is 730?

Solution:

2, 4, 10, 28, 82, 244, 730. 730 is the 7th term.

Question 4: Which term of GP 2, 6, 18, … is 4374?

Solution:

2×3ⁿ⁻¹ = 4374 → 3ⁿ⁻¹ = 2187 = 3⁷ → n = 8

Explicit: tₙ = 2×3ⁿ⁻¹. Recursive: t₁=2, tₙ=3tₙ₋₁

Question 5: Ball dropped from 80 m, bouncing to 60% height

Solution:

(i) After 5th bounce: 80×(0.6)⁵ ≈ 6.22 m

(ii) Total distance by 6th hit ≈ 80 + 2×(48+28.8+17.28+10.368+6.2208) ≈ 301.34 m

Question 6: Which term of 2, 2√2, 4, … is 128?

Solution:

r = √2. tₙ = 2^((n+1)/2) = 128 = 2⁷ → n = 13

Question 7: Sierpinski Square Carpet

Solution:

(i) Red squares: 1, 8, 64, 512 (×8 each stage)

(ii) Stage 4 = 4096; Stage 5 = 32768

(iii) Explicit: tₙ = 8ⁿ. Recursive: t₀=1, tₙ=8tₙ₋₁

(iv) Area: sₙ = (8/9)ⁿ → approaches 0

End-of-Chapter Exercises — Solutions

Question 1: 31st term of AP — 11th term=38, 16th term=73

Solution:

5d = 35 → d=7; a=−32. t₃₁ = −32+30×7 = 178

Question 2: AP — 3rd term=16, 7th term exceeds 5th by 12

Solution:

2d=12 → d=6; a=4. AP: 4, 10, 16, 22, 28, …

Question 3: Three-digit numbers divisible by 7

Solution:

AP: 105, 112, …, 994. n = 128

Question 4: Multiples of 4 between 10 and 250

Solution:

AP: 12, 16, …, 248. n = 60

Question 5: GP — sum of first two terms = −4, 5th term = 4 × 3rd term

Solution:

r² = 4 → r = ±2

r=2: a=−4/3. GP: −4/3, −8/3, −16/3, …

r=−2: a=4. GP: 4, −8, 16, −32, …

Question 6: Express 100 as sum of consecutive natural numbers

Solution:

18+19+20+21+22 = 100 (5 terms)

9+10+11+12+13+14+15+16 = 100 (8 terms)

Question 7: Bacteria doubling every hour, starting with 30

Solution:

Bₙ = 30×2ⁿ. After 2nd hour: 120. After 4th hour: 480.

Question 8: AP — t₄+t₈=24, t₆+t₁₀=44

Solution:

d=5; a=−13. First three terms: −13, −8, −3

Question 9: Smallest n where 1+2+…+n > 1000

Solution:

n(n+1)/2 > 1000. n=44 gives 990; n=45 gives 1035. Answer: n = 45

Question 10: Which term of GP 2, 8, 32, … is 131072?

Solution:

a=2, r=4. tₙ = 2²ⁿ⁻¹ = 2¹⁷ → n = 9

Question 11: GP — sum of first 3 terms = 13/12, product = −1

Solution:

a = −1. 12r² + 25r + 12 = 0 → r = −4/3 or −3/4

Terms: 3/4, −1, 4/3 or 4/3, −1, 3/4

Question 12: If 4th, 10th, 16th terms of GP are x, y, z — prove x, y, z in GP

Solution:

x=ar³, y=ar⁹, z=ar¹⁵. y/x = r⁶ = z/y. ∴ x, y, z are in GP. (Proved)

Question 13: GP — sum of first 3 terms = 26, sum of squares = 364

Solution:

r=3 or r=1/3. Terms: 2, 6, 18 or 18, 6, 2

Question 14: P₁=1, P₂=2, Pₙ = P₁+…+Pₙ₋₁+1 for n>2

Solution:

1, 2, 4, 8, 16, 32, 64, 128. Explicit: Pₙ = 2ⁿ⁻¹. Recursive: Pₙ = 2Pₙ₋₁ (n≥3)

Question 15: W₁=1, W₂=2, Wₙ = Wₙ₋₁+Wₙ₋₂+2 for n>2

Solution:

W₁=1; W₂=2; W₃=5; W₄=9; W₅=16; W₆=27; W₇=45; W₈=74

Happy Learning! Keep exploring the beautiful patterns in mathematics.