Sequences & Series | Class 11 Maths

What is a Sequence?

📌 Definition

A sequence is an arrangement of numbers in a definite order according to some rule. It can be viewed as a function whose domain is the set of natural numbers (or some subset of it).

Each number in a sequence is called a term.
Terms are denoted by a₁, a₂, a₃, …, aₙ — where n is the position.
The nth term (aₙ) is also called the general term.
A finite sequence has a fixed number of terms. An infinite sequence never ends.

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IMAGE SPACE — Sequence examples diagram

Add an image showing: 2, 4, 8, 16, 32 (ancestors) as a visual timeline

🔢 Types of Sequences

Finite: Sequence of ancestors over 10 generations → a₁=2, a₂=4, …, a₁₀=1024
Infinite: Successive quotients of 10÷3 → 3, 3.3, 3.33, 3.333, …
Fibonacci: a₁=a₂=1, aₙ=aₙ₋₁+aₙ₋₂ → 1, 1, 2, 3, 5, 8, 13, …
Even Numbers: General term aₙ = 2n → 2, 4, 6, 8, …
Odd Numbers: General term aₙ = 2n–1 → 1, 3, 5, 7, …

∑ A Series is the sum of terms of a sequence: a₁ + a₂ + a₃ + … + aₙ. It is written in compact sigma (Σ) notation.

📝 MCQ Practice

Sequences — Test Yourself

1 If aₙ = 4n – 3, find the value of a₁₇.

A 63
B 68
C 65
D 61

Solution: aₙ = 4n – 3
a₁₇ = 4(17) – 3 = 68 – 3 = 65 ✓

2 The 20th term of the sequence defined by aₙ = (n–1)(2–n)(3+n) is:

A –7800
B –7866
C 7866
D –7654

Solution: aₙ = (n–1)(2–n)(3+n)
a₂₀ = (20–1)(2–20)(3+20) = 19 × (–18) × 23
= 19 × (–414) = –7866 ✓

3 For the Fibonacci sequence where a₁ = a₂ = 1 and aₙ = aₙ₋₁ + aₙ₋₂, what is a₆?

A 7
B 9
C 8
D 13

Solution: Fibonacci: 1, 1, 2, 3, 5, 8, 13, …
a₁=1, a₂=1, a₃=2, a₄=3, a₅=5, a₆ = a₄ + a₅ = 3 + 5 = 8 ✓

Geometric Progression (G.P.)

📌 Definition

A sequence a₁, a₂, a₃, … is called a Geometric Progression (G.P.) if each term is non-zero and the ratio of any term to its preceding term is constant.

This constant is called the Common Ratio (r): r = aₖ₊₁ / aₖ
General form of G.P.: a, ar, ar², ar³, … where a = first term
The nth term of G.P.: aₙ = arⁿ⁻¹

🎯 Key Formulas — G.P.

aₙ = arⁿ⁻¹

Sₙ = a(rⁿ – 1) / (r – 1) [when r ≠ 1]

Sₙ = na [when r = 1]

where a = first term, r = common ratio, n = number of terms

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IMAGE SPACE — G.P. visual

Add an image: arrows showing ×r multiplication between terms a → ar → ar² → ar³

💡 Examples of G.P.

2, 4, 8, 16, 32, … → r = 2 (doubling)
1/9, –1/27, 1/81, –1/243, … → r = –1/3
0.01, 0.0001, 0.000001, … → r = 0.01
Ancestors: 2, 4, 8, … 1024 over 10 generations → G.P. with a=2, r=2, n=10

📝 MCQ Practice

Geometric Progression — Test Yourself

4 Find the 10th term of the G.P. 5, 25, 125, …

A 5⁹
B 5¹⁰
C 5⁸
D 5¹¹

Solution: Here a = 5, r = 25/5 = 5
a₁₀ = ar⁹ = 5 × 5⁹ = 5¹⁰ ✓

5 In a G.P., the 3rd term is 24 and the 6th term is 192. What is the 10th term?

A 1536
B 6144
C 3072
D 2048

Solution: a₃ = ar² = 24 … (1) | a₆ = ar⁵ = 192 … (2)
Dividing (2)÷(1): r³ = 8 → r = 2
From (1): a(4) = 24 → a = 6
a₁₀ = 6 × 2⁹ = 6 × 512 = 3072 ✓

6 How many terms of G.P. 3, 3/2, 3/4, … are needed to give the sum 3069/512?

A 8
B 9
C 10
D 12

Solution: a = 3, r = 1/2, Sₙ = 3069/512
Sₙ = a(1–rⁿ)/(1–r) = 3(1–(1/2)ⁿ)/(1/2) = 6(1–1/2ⁿ)
3069/512 = 6(1–1/2ⁿ) → 3069/3072 = 1–1/2ⁿ
1/2ⁿ = 3/3072 = 1/1024 = 1/2¹⁰ → n = 10 ✓

Geometric Mean (G.M.)

📌 Definition

The Geometric Mean (G) of two positive numbers a and b is the number G = √(ab). Notice that a, G, b form a G.P. since G/a = b/G = √(b/a).

G.M. of 2 and 8 = √(2×8) = √16 = 4 → sequence 2, 4, 8 is a G.P.
To insert n G.M.s between a and b: common ratio r = (b/a)^(1/(n+1))
G₁ = a(b/a)^(1/(n+1)), G₂ = a(b/a)^(2/(n+1)), …, Gₙ = a(b/a)^(n/(n+1))

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IMAGE SPACE — G.M. insertion diagram

Add an image showing: 1, G₁, G₂, G₃, 256 as a number line with G.P. labels

🎯 G.M. Formula

G = √(ab)

For n G.M.s between a and b:
r = (b/a)^(1/(n+1)) | Gₖ = a · r^k

📝 MCQ Practice

Geometric Mean — Test Yourself

7 Three numbers are inserted between 1 and 256 to form a G.P. The three inserted numbers (for positive r) are:

A 2, 8, 32
B 4, 16, 64
C 8, 32, 128
D 2, 16, 128

Solution: 1, G₁, G₂, G₃, 256 is a G.P.
256 = 1 × r⁴ → r⁴ = 256 → r = 4 (positive root)
G₁ = 1×4 = 4, G₂ = 1×16 = 16, G₃ = 1×64 = 64
Answer: 4, 16, 64 ✓

8 The geometric mean of two numbers a and b is 8 and their arithmetic mean is 10. The value of |a – b| is:

A 8
B 10
C 12
D 14

Solution: A.M. = 10 → a+b = 20; G.M. = 8 → ab = 64
(a–b)² = (a+b)² – 4ab = 400 – 256 = 144
|a – b| = 12 ✓ (Numbers are 4 and 16)

9 If the G.M. of two numbers is 6 and one of them is 4, what is the other number?

A 8
B 9
C 36
D 3

Solution: G = √(ab) = 6 → ab = 36
If a = 4 → b = 36/4 = 9 ✓
Verify: √(4 × 9) = √36 = 6 ✓

Relationship Between A.M. and G.M.

📌 The A.M.–G.M. Inequality

For any two positive real numbers a and b:

A.M. (Arithmetic Mean) = (a + b) / 2
G.M. (Geometric Mean) = √(ab)
The fundamental inequality: A.M. ≥ G.M.
Equality holds when and only when a = b.

🎯 A.M.–G.M. Proof

A – G = (a+b)/2 – √(ab) = (√a – √b)² / 2 ≥ 0

Therefore: A.M. ≥ G.M. always holds for positive numbers
Equality iff a = b (i.e., √a = √b)

🔵 When A.M. = G.M.

This happens only when the two numbers are equal: a = b. The G.P. a, G, b becomes a, a, a — a constant sequence.

🔴 When A.M. > G.M.

For any two distinct positive numbers, A.M. is strictly greater than G.M. This is widely used in optimization problems.

🟣 Finding numbers from A.M. & G.M.

If A.M. = A and G.M. = G, then a + b = 2A and ab = G². Solve the quadratic: x² – 2Ax + G² = 0.

📊 Real Application

If A.M. = 10 and G.M. = 8: numbers are 4 and 16. Check: (4+16)/2=10 ✓ and √(4×16)=8 ✓

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IMAGE SPACE — A.M. vs G.M. comparison

Add a diagram: number line showing A.M. ≥ G.M. with (√a – √b)² ≥ 0 geometric interpretation

📝 MCQ Practice

A.M. & G.M. — Test Yourself

10 If A.M. and G.M. of two numbers are 10 and 8, find the two numbers.

A 2 and 18
B 4 and 16
C 6 and 14
D 8 and 12

Solution: A.M. = 10 → a+b = 20; G.M. = 8 → ab = 64
x² – 20x + 64 = 0 → (x–4)(x–16) = 0
a = 4, b = 16 ✓

11 For two positive numbers a and b with a ≠ b, which statement is always true?

A A.M. = G.M.
B A.M. > G.M.
C G.M. > A.M.
D A.M. + G.M. = 0

Solution: A – G = (√a – √b)² / 2
Since a ≠ b → (√a – √b)² > 0 → A – G > 0
Therefore A.M. > G.M. strictly when a ≠ b ✓

12 The number of bacteria in a culture doubles every hour. Starting with 30 bacteria, how many will be there at the end of the 4th hour?

A 240
B 300
C 480
D 960

Solution: G.P. with a = 30, r = 2
After 4th hour → a₅ (term at position 5) = ar⁴ = 30 × 2⁴ = 30 × 16 = 480 ✓

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Quick Summary

🗂️ Chapter 8 at a Glance

Sequence: Ordered list of numbers by a rule. Terms: a₁, a₂, …, aₙ. General term = aₙ.
Series: Sum of sequence terms. Written as Σ (sigma) notation.
G.P.: Constant ratio r between consecutive terms. aₙ = arⁿ⁻¹.
Sum of G.P.: Sₙ = a(rⁿ–1)/(r–1) for r≠1; Sₙ = na for r=1.
G.M.: G = √(ab) between positive numbers a and b.
A.M.–G.M. Inequality: (a+b)/2 ≥ √(ab), equality iff a=b.

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Sequences &Series

📌 Definition

🔢 Types of Sequences

Sequences — Test Yourself

📌 Definition

💡 Examples of G.P.

Geometric Progression — Test Yourself

📌 Definition

Geometric Mean — Test Yourself

📌 The A.M.–G.M. Inequality

🔵 When A.M. = G.M.

🔴 When A.M. > G.M.

🟣 Finding numbers from A.M. & G.M.

📊 Real Application

A.M. & G.M. — Test Yourself

🗂️ Chapter 8 at a Glance

Sequences &
Series