Class 9 • Mathematics • Ganita Manjari
Chapter 8: Predicting What Comes Next
Exploring Sequences & Progressions
Complete study notes covering Sequences, Explicit & Recursive Rules, AP, GP, and the Virahānka–Fibonacci sequence — with examples, formulas, and exam questions.
📐 Sequences
➕ Arithmetic Progression
✖️ Geometric Progression
🔁 Recursive Rules
🔢 Sum of Natural Numbers
🔺 Fractals
➕ Arithmetic Progression
✖️ Geometric Progression
🔁 Recursive Rules
🔢 Sum of Natural Numbers
🔺 Fractals
📋 Contents
8.1 Introduction to Sequences
We see patterns everywhere — in nature, art, music, and finance. In mathematics, a sequence is an ordered list of numbers (or terms) arranged in a particular order following a specific rule.
📌 Common Sequences You Already Know
| Sequence Name | Terms | Pattern |
|---|---|---|
| Natural Numbers | 1, 2, 3, 4, 5, 6, … | Add 1 each time |
| Odd Numbers | 1, 3, 5, 7, 9, 11, … | Add 2 each time |
| Triangular Numbers | 1, 3, 6, 10, 15, 21, … | Sum of 1+2+3+… up to n |
| Square Numbers | 1, 4, 9, 16, 25, 36, … | 1², 2², 3², n² |
Key Definition
A sequence is an ordered list of numbers. Each number is called a term. Sequences may be finite (e.g., 6, 12, 24, 48, 96) or infinite (indicated by …).
A sequence is an ordered list of numbers. Each number is called a term. Sequences may be finite (e.g., 6, 12, 24, 48, 96) or infinite (indicated by …).
📝 Term Notation
We use subscript notation: t₁ = first term, t₂ = second term, tₙ = nth term.
Example: For odd numbers, t₁ = 1, t₂ = 3, t₃ = 5, t₄ = 7 …
Triangular Numbers Trick
Each triangular number is the sum of natural numbers up to that term. The 5th triangular number = 1 + 2 + 3 + 4 + 5 = 15. Square numbers = sum of consecutive odd numbers: 9 = 1 + 3 + 5.
Each triangular number is the sum of natural numbers up to that term. The 5th triangular number = 1 + 2 + 3 + 4 + 5 = 15. Square numbers = sum of consecutive odd numbers: 9 = 1 + 3 + 5.
8.2 Explicit Rule for a Sequence
An explicit formula uses the term’s position number n to directly calculate the value of that term — without needing to know previous terms.
tₙ = f(n) → Substitute n = 1, 2, 3, … to get terms directly
📘 Example 1: Odd Numbers
uₙ = 2n − 1
Substituting: u₁ = 2(1)−1 = 1, u₂ = 2(2)−1 = 3, u₃ = 2(3)−1 = 5 ✓
Why is an explicit formula useful?
You can find the 20th, 53rd, or 1000th term directly without computing all previous terms! For example, the 53rd odd number = 2(53)−1 = 105.
You can find the 20th, 53rd, or 1000th term directly without computing all previous terms! For example, the 53rd odd number = 2(53)−1 = 105.
📘 Example 2: Sequence sₙ = 5n − 2
First 6 terms: s₁=3, s₂=8, s₃=13, s₄=18, s₅=23, s₆=28
Is 308 a term? Solve 5n − 2 = 308 → 5n = 310 → n = 62. ✅ Yes, it is the 62nd term.
Is 471 a term? Solve 5n − 2 = 471 → 5n = 473 → n = 94.6. ❌ Not a natural number, so 471 is NOT a term.
Important Rule
The subscript n must always be a positive integer (1, 2, 3, …). If solving gives a decimal or negative value, the number is NOT a term of that sequence.
The subscript n must always be a positive integer (1, 2, 3, …). If solving gives a decimal or negative value, the number is NOT a term of that sequence.
🔢 Exercise: tn = 3n − 7
| n | 1 | 2 | 3 | 12 | 18 | 50 |
|---|---|---|---|---|---|---|
| tₙ | −4 | −1 | 2 | 29 | 47 | 143 |
Is 332 a term? 3n − 7 = 332 → n = 113. ✅ Yes, 113th term. | Is 557 a term? 3n − 7 = 557 → n = 188. ✅ Yes!
8.3 Recursive Rule for a Sequence
A recursive formula defines each term using the previous term(s). You need to know earlier terms to find the next ones.
t₁ = first term, tₙ = tₙ₋₁ + d (for n ≥ 2)
📘 Example: Sequence 1, 4, 7, 10, 13, …
Explicit Ruletₙ = 3n − 2
Recursive Rulet₁ = 1, tₙ = tₙ₋₁ + 3 for n ≥ 2
📘 Example 3: u₁ = 1, uₙ = 2uₙ₋₁ + 3
u₁ = 1
u₂ = 2 × 1 + 3 = 5
u₃ = 2 × 5 + 3 = 13
u₄ = 2 × 13 + 3 = 29
Sequence: 1, 5, 13, 29, …
u₂ = 2 × 1 + 3 = 5
u₃ = 2 × 5 + 3 = 13
u₄ = 2 × 13 + 3 = 29
Sequence: 1, 5, 13, 29, …
📘 Example 4: s₁ = 3, sₙ = sₙ₋₁(sₙ₋₁ − 1)
s₁ = 3
s₂ = 3 × (3 − 1) = 6
s₃ = 6 × (6 − 1) = 30
s₄ = 30 × (30 − 1) = 870
Sequence: 3, 6, 30, 870, …
s₂ = 3 × (3 − 1) = 6
s₃ = 6 × (6 − 1) = 30
s₄ = 30 × (30 − 1) = 870
Sequence: 3, 6, 30, 870, …
Common Mistake
In a recursive formula, you MUST know the first term (or first few terms) to use the rule. Without the starting value, you cannot compute any other term!
In a recursive formula, you MUST know the first term (or first few terms) to use the rule. Without the starting value, you cannot compute any other term!
Virahānka–Fibonacci Sequence
This is the most famous sequence involving a recursive rule with two previous terms:
V₁ = 1, V₂ = 2, Vₙ = Vₙ₋₁ + Vₙ₋₂ for n ≥ 3
V₃ = 2 + 1 = 3
V₄ = 3 + 2 = 5
V₅ = 5 + 3 = 8
Sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
V₄ = 3 + 2 = 5
V₅ = 5 + 3 = 8
Sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
Historical Context
This sequence was first written down and studied by the Indian scholar Virahānka in his work Vṛttajātisamuchaya in the 7th century CE — in the context of Prakrit poetry and metre! It was further studied by Gopāla (c. 1135 CE) and Hemachandra (c. 1150 CE), and later by the Italian mathematician Fibonacci (c. 1200 CE). India did it first! 🇮🇳
This sequence was first written down and studied by the Indian scholar Virahānka in his work Vṛttajātisamuchaya in the 7th century CE — in the context of Prakrit poetry and metre! It was further studied by Gopāla (c. 1135 CE) and Hemachandra (c. 1150 CE), and later by the Italian mathematician Fibonacci (c. 1200 CE). India did it first! 🇮🇳
In Nature
The Virahānka–Fibonacci sequence appears in the spiral arrangement of seeds in sunflowers, the number of petals in flowers, the branching of trees, and many other natural phenomena. It plays an important role throughout mathematics and science.
The Virahānka–Fibonacci sequence appears in the spiral arrangement of seeds in sunflowers, the number of petals in flowers, the branching of trees, and many other natural phenomena. It plays an important role throughout mathematics and science.
8.4 Arithmetic Progressions (AP)
An Arithmetic Progression (AP) is a sequence in which the difference between any two consecutive terms is always constant. This constant is called the common difference (d).
General AP: a, a+d, a+2d, a+3d, …, a+(n−1)d
nth term: tₙ = a + (n − 1) × d
First Term (a)The starting value of the AP
Common Difference (d)d = any term − previous term (can be +ve, −ve, or 0)
📌 Examples of APs
| Sequence | First Term (a) | Common Diff. (d) | nth term formula |
|---|---|---|---|
| 1, 5, 9, 13, 17, … | 1 | 4 | 4n − 3 |
| 1, 4, 7, 10, … | 1 | 3 | 3n − 2 |
| 11, 7, 3, −1, −5, … | 11 | −4 | 15 − 4n |
| 2, 5, 8, 11, … | 2 | 3 | 3n − 1 |
📘 Recursive Rule for an AP
t₁ = a, tₙ = tₙ₋₁ + d for n ≥ 2
📘 AP on a Graph
Key Visual Property
When you plot the pairs (n, tₙ) for an AP, the points always lie on a straight line! This is because the nth term formula is linear (degree 1 in n). For GP, the points form a curve.
When you plot the pairs (n, tₙ) for an AP, the points always lie on a straight line! This is because the nth term formula is linear (degree 1 in n). For GP, the points form a curve.
📘 Real-Life Example — Taxi Fare (Example 5)
Fixed booking fee: ₹200 | Rate: ₹40 per km
After 1 km: ₹200 + ₹40 = ₹240
After 2 km: ₹200 + ₹80 = ₹280
After 3 km: ₹200 + ₹120 = ₹320
AP: 240, 280, 320, 360, … (a=240, d=40)
nth term: tₙ = 200 + 40n (n = km travelled)
After 10 km: t₁₀ = 200 + 400 = ₹600
After 2 km: ₹200 + ₹80 = ₹280
After 3 km: ₹200 + ₹120 = ₹320
AP: 240, 280, 320, 360, … (a=240, d=40)
nth term: tₙ = 200 + 40n (n = km travelled)
After 10 km: t₁₀ = 200 + 400 = ₹600
AP Quick Test
Subtract any term from the next one. If the difference is the same throughout, it’s an AP. Example: 2, 5, 8, 11 → 5−2=3, 8−5=3, 11−8=3. ✅ AP with d=3.
Subtract any term from the next one. If the difference is the same throughout, it’s an AP. Example: 2, 5, 8, 11 → 5−2=3, 8−5=3, 11−8=3. ✅ AP with d=3.
8.5 Sum of the First n Natural Numbers
How to find the sum 1 + 2 + 3 + … + n without adding them all up?
🔢 The Clever Method (Gauss’s Trick)
S = 1 + 2 + 3 + … + 9 + 10
S = 10 + 9 + 8 + … + 2 + 1
――――――――――――――――――――――――――――――――
2S = 11 + 11 + 11 + … + 11 (10 times)
2S = 110 → S = 55
S = 10 + 9 + 8 + … + 2 + 1
――――――――――――――――――――――――――――――――
2S = 11 + 11 + 11 + … + 11 (10 times)
2S = 110 → S = 55
Sₙ = 1 + 2 + 3 + … + n = n(n + 1) / 2
Historical Note — Āryabhaṭa
The first known written record of this formula appears in Āryabhaṭa’s Āryabhaṭīya (Chapter 2, Verse 19). He described that the sum equals: (first term + last term) ÷ 2, multiplied by the number of terms — which is exactly the average of first and last, times n.
The first known written record of this formula appears in Āryabhaṭa’s Āryabhaṭīya (Chapter 2, Verse 19). He described that the sum equals: (first term + last term) ÷ 2, multiplied by the number of terms — which is exactly the average of first and last, times n.
📌 Using the Formula
| Sum | Formula | Answer |
|---|---|---|
| S₁₀ | 10 × 11 / 2 | 55 |
| S₂₀ | 20 × 21 / 2 | 210 |
| S₅₀ | 50 × 51 / 2 | 1275 |
| S₁₀₀ | 100 × 101 / 2 | 5050 |
📘 Sum of Consecutive Numbers
To find 25 + 26 + 27 + … + 58:
= (1+2+…+58) − (1+2+…+24)
= S₅₈ − S₂₄
= (58×59)/2 − (24×25)/2
= 1711 − 300
= 1411
= S₅₈ − S₂₄
= (58×59)/2 − (24×25)/2
= 1711 − 300
= 1411
Triangular Numbers Connection
The nth triangular number = Sₙ = n(n+1)/2. So the 10th triangular number = 10×11/2 = 55, the 17th = 17×18/2 = 153, the 80th = 80×81/2 = 3240.
The nth triangular number = Sₙ = n(n+1)/2. So the 10th triangular number = 10×11/2 = 55, the 17th = 17×18/2 = 153, the 80th = 80×81/2 = 3240.
8.6 Geometric Progressions (GP)
A Geometric Progression (GP) is a sequence in which each term is obtained by multiplying the previous term by a fixed constant. This constant is called the common ratio (r).
General GP: a, ar, ar², ar³, …, arⁿ⁻¹
nth term: tₙ = a × rⁿ⁻¹
First Term (a)The starting value of the GP
Common Ratio (r)r = any term ÷ previous term
📌 Examples of GPs
| Sequence | a | r | nth term |
|---|---|---|---|
| 3, 6, 12, 24, 48, … | 3 | 2 | 3 × 2ⁿ⁻¹ |
| 1, 2, 4, 8, 16, … | 1 | 2 | 2ⁿ⁻¹ |
| 1, 3, 9, 27, 81, … | 1 | 3 | 3ⁿ⁻¹ |
| 1, −1, 1, −1, … | 1 | −1 | (−1)ⁿ⁻¹ |
| 5, 15/4, 45/16, … | 5 | 3/4 | 5 × (3/4)ⁿ⁻¹ |
📘 Recursive Rule for a GP
t₁ = a, tₙ = r × tₙ₋₁ for n ≥ 2
📘 GP vs AP on a Graph
Visual Difference
AP points (n, tₙ) lie on a straight line. GP points lie on a curve (exponential shape). When r > 1, the curve rises steeply (exponential growth). When 0 < r < 1, it decreases toward 0 (exponential decay).
AP points (n, tₙ) lie on a straight line. GP points lie on a curve (exponential shape). When r > 1, the curve rises steeply (exponential growth). When 0 < r < 1, it decreases toward 0 (exponential decay).
📘 Real-Life Example — Bouncing Ball (Example 10)
Ball dropped from 24 feet. Each bounce reaches ¾ (75%) of previous height.
1st bounce: 24 × 0.75 = 18.000 ft
2nd bounce: 18 × 0.75 = 13.500 ft
3rd bounce: 13.5 × 0.75 = 10.125 ft
4th bounce: 10.125 × 0.75 = 7.594 ft
5th bounce: 7.594 × 0.75 = 5.695 ft
GP: a = 18, r = 3/4
tₙ = 18 × (3/4)ⁿ⁻¹
1/6 of 24 = 4 ft → after 7th bounce: 3.20 ft < 4 ft ✓
2nd bounce: 18 × 0.75 = 13.500 ft
3rd bounce: 13.5 × 0.75 = 10.125 ft
4th bounce: 10.125 × 0.75 = 7.594 ft
5th bounce: 7.594 × 0.75 = 5.695 ft
GP: a = 18, r = 3/4
tₙ = 18 × (3/4)ⁿ⁻¹
1/6 of 24 = 4 ft → after 7th bounce: 3.20 ft < 4 ft ✓
GP Quick Test
Divide any term by the previous term. If the ratio is always the same, it’s a GP. Example: 3, 6, 12, 24 → 6/3=2, 12/6=2, 24/12=2. ✅ GP with r=2.
Divide any term by the previous term. If the ratio is always the same, it’s a GP. Example: 3, 6, 12, 24 → 6/3=2, 12/6=2, 24/12=2. ✅ GP with r=2.
8.6.1 Fractals and Geometric Progressions
Fractals are shapes that repeat themselves at every scale — zoom in and you see the same pattern repeating infinitely. They are created using simple recursive rules but produce incredibly complex beautiful designs.
Sierpiński Triangle
Start with an equilateral triangle (Stage 0). Join midpoints of all 3 sides → 4 smaller triangles → remove the central one. Repeat forever! Each black triangle is replaced by 3 smaller triangles in the next stage.
Start with an equilateral triangle (Stage 0). Join midpoints of all 3 sides → 4 smaller triangles → remove the central one. Repeat forever! Each black triangle is replaced by 3 smaller triangles in the next stage.
📌 Sierpiński Triangle — GP Patterns
| Stage (n) | 0 | 1 | 2 | 3 | 4 | n |
|---|---|---|---|---|---|---|
| Black triangles (tₙ) | 1=3⁰ | 3=3¹ | 9=3² | 27=3³ | 81=3⁴ | 3ⁿ |
| Shaded area (sₙ) | 1 | 3/4 | (3/4)² | (3/4)³ | (3/4)⁴ | (3/4)ⁿ |
Number of trianglesGP with a=1, r=3
tₙ = 3ⁿ (grows very fast!)
tₙ = 3ⁿ (grows very fast!)
Area of black regionGP with a=1, r=3/4
sₙ = (3/4)ⁿ (approaches 0!)
sₙ = (3/4)ⁿ (approaches 0!)
Fractals in Nature
Fractals appear in: branching of trees, cauliflower and broccoli florets, snowflakes, coastlines, river networks, and even our lungs! The beautiful complexity comes from repeating a simple rule infinitely.
Fractals appear in: branching of trees, cauliflower and broccoli florets, snowflakes, coastlines, river networks, and even our lungs! The beautiful complexity comes from repeating a simple rule infinitely.
Chapter Summary — All Key Formulas at a Glance
Sequence
An ordered list of numbers. Each number = a term. Can be finite or infinite.
An ordered list of numbers. Each number = a term. Can be finite or infinite.
Explicit Formula
Calculates tₙ directly using n.
Example: tₙ = 2n−1 (odd numbers)
Calculates tₙ directly using n.
Example: tₙ = 2n−1 (odd numbers)
Recursive Formula
Uses previous terms.
Example: t₁=1, tₙ=tₙ₋₁+3
Uses previous terms.
Example: t₁=1, tₙ=tₙ₋₁+3
Virahānka–Fibonacci
V₁=1, V₂=2, Vₙ=Vₙ₋₁+Vₙ₋₂
Sequence: 1,2,3,5,8,13,21,…
V₁=1, V₂=2, Vₙ=Vₙ₋₁+Vₙ₋₂
Sequence: 1,2,3,5,8,13,21,…
Arithmetic Progression (AP)
tₙ = a + (n−1)d
Recursive: tₙ = tₙ₋₁ + d
Points lie on a straight line.
tₙ = a + (n−1)d
Recursive: tₙ = tₙ₋₁ + d
Points lie on a straight line.
Sum of n Natural Numbers
Sₙ = n(n+1)/2
= 1+2+3+…+n
Also = nth triangular number
Sₙ = n(n+1)/2
= 1+2+3+…+n
Also = nth triangular number
Geometric Progression (GP)
tₙ = arⁿ⁻¹
Recursive: tₙ = r·tₙ₋₁
Points form a curve.
tₙ = arⁿ⁻¹
Recursive: tₙ = r·tₙ₋₁
Points form a curve.
Fractals → GP
Sierpiński triangle:
triangles = 3ⁿ, area = (3/4)ⁿ
Sierpiński triangle:
triangles = 3ⁿ, area = (3/4)ⁿ
🗝️ AP vs GP — Side by Side
| Feature | AP | GP |
|---|---|---|
| Operation | Add/subtract constant d | Multiply by constant r |
| nth term | a + (n−1)d | arⁿ⁻¹ |
| Recursive | tₙ = tₙ₋₁ + d | tₙ = r·tₙ₋₁ |
| Graph shape | Straight line | Curve (exponential) |
| Example | 3, 7, 11, 15, … (d=4) | 2, 6, 18, 54, … (r=3) |
Exam Practice Questions with Solutions
🟢 Exercise Set 8.1 — Sequences & Explicit/Recursive Rules
Q1. Find the 10th and 15th terms of the sequence tₙ = 5n − 3.
t₁₀ = 5(10) − 3 = 47 | t₁₅ = 5(15) − 3 = 72
Q2. Is 97 a term of the sequence tₙ = 5n − 3?
5n − 3 = 97 → 5n = 100 → n = 20. ✅ Yes! 97 is the 20th term.
Q3. A sequence is given by t₁ = −5, tₙ₊₁ = tₙ + 3 for n ≥ 1. Find the first 5 terms. Is 52 a term?
Terms: −5, −2, 1, 4, 7. | General term: tₙ = 3n − 8. For 52: 3n − 8 = 52 → n = 20. ✅ Yes, 52 is the 20th term.
🟡 Exercise Set 8.2 — Arithmetic Progressions
Q4. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, …
a = 3, d = 5. t₁₀ = 3 + 9(5) = 48. t₂₆ = 3 + 25(5) = 128.
Q5. Which term of the AP: 21, 18, 15, … is −81? Also, is 0 a term?
a=21, d=−3. tₙ=−81: 21+(n−1)(−3)=−81 → n=35. 35th term. For 0: 21−3(n−1)=0 → n=8. ✅ 0 is the 8th term.
Q6. An AP has 50 terms. The 3rd term is 12 and the last term is 106. Find the 29th term.
a + 2d = 12 and a + 49d = 106. Subtracting: 47d = 94 → d = 2, a = 8. t₂₉ = 8 + 28(2) = 64.
Q7. How many 2-digit numbers are divisible by 3?
AP: 12, 15, 18, …, 99. a=12, d=3. tₙ=99 → 12+3(n−1)=99 → n=30. There are 30 two-digit multiples of 3.
🔴 Exercise Set 8.3 — Geometric Progressions
Q8. Find the 10th and nth terms of the GP: 5, 25, 125, …
a=5, r=5. tₙ = 5 × 5ⁿ⁻¹ = 5ⁿ. t₁₀ = 5¹⁰ = 9,765,625.
Q9. Which term of the GP: 2, 6, 18, … is 4374? Write explicit and recursive formulas.
a=2, r=3. tₙ = 2 × 3ⁿ⁻¹ = 4374 → 3ⁿ⁻¹ = 2187 = 3⁷ → n=8. It’s the 8th term. Recursive: t₁=2, tₙ=3tₙ₋₁.
Q10. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
t₈ = a × 2⁷ = 192 → a = 192/128 = 1.5. t₁₂ = 1.5 × 2¹¹ = 1.5 × 2048 = 3072.
⭐ End-of-Chapter Challenge Questions
Q11. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
t₁₁ = a+10d = 38 and t₁₆ = a+15d = 73. Subtracting: 5d = 35 → d = 7, a = −32. t₃₁ = −32+30(7) = 178.
Q12. The number of bacteria doubles every hour. If there were 30 originally, how many after the nth hour?
GP with a=30, r=2. After nth hour: tₙ₊₁ = 30 × 2ⁿ. After 2nd hr: 120. After 4th hr: 480. After nth hr: 30 × 2ⁿ.
Q13. Find the smallest n such that 1+2+3+…+n > 1000.
n(n+1)/2 > 1000 → n(n+1) > 2000. Try n=44: 44×45=1980 (no). n=45: 45×46=2070 > 2000. ✅ n = 45.
🌟
Fun Fact: Gauss and the Sum Formula
Legend says the young Carl Friedrich Gauss (age 10) surprised his teacher by instantly adding 1 to 100. He mentally paired 1+100=101, 2+99=101, … giving 50 pairs × 101 = 5050. This is essentially the same trick as the formula Sₙ = n(n+1)/2, which Āryabhaṭa had already documented over 1000 years earlier!
Legend says the young Carl Friedrich Gauss (age 10) surprised his teacher by instantly adding 1 to 100. He mentally paired 1+100=101, 2+99=101, … giving 50 pairs × 101 = 5050. This is essentially the same trick as the formula Sₙ = n(n+1)/2, which Āryabhaṭa had already documented over 1000 years earlier!
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