Sequences and Series — JEE Main Mathematics Notes & Formulas

1. Chapter at a Glance

A sequence is an ordered list of numbers that follows a rule; a series is the sum of its terms. The two stars of this chapter are the Arithmetic Progression (AP), where we add a fixed number each time, and the Geometric Progression (GP), where we multiply by a fixed number. JEE Main asks about sums of AP/GP, the AM–GM inequality for maximum or minimum values, and special series (Σn², telescoping sums).

What JEE asks most

Finding a term or a sum from given information about an AP or GP.
Minimum or maximum value using AM ≥ GM.
Sums like Σn(n + 1) or Σ1/(n(n + 1)) using standard formulas and the method of differences.
Infinite GP and arithmetico-geometric series.

2. Arithmetic Progression (AP)

An AP has a constant common difference d: a, a + d, a + 2d, …

a_n = a + (n − 1)d, S_n = (n/2)[2a + (n − 1)d] = (n/2)(first + last)

a_n = S_n − S_n−1 for any sequence.
If S_n is a quadratic in n with no constant term, the sequence is an AP and d = 2 × (coefficient of n²).
Three numbers in AP are best taken as a − d, a, a + d; four as a − 3d, a − d, a + d, a + 3d.
a, b, c are in AP ⇔ 2b = a + c. The arithmetic mean of a and b is (a + b)/2.
Adding, subtracting, multiplying or dividing every term by a constant keeps an AP an AP.

3. Geometric Progression (GP)

A GP has a constant common ratio r: a, ar, ar², …

a_n = arⁿ⁻¹, S_n = a(rⁿ − 1)/(r − 1) (r ≠ 1), S_∞ = a/(1 − r) for |r| < 1

Three numbers in GP are best taken as a/r, a, ar.
a, b, c are in GP ⇔ b² = ac. The geometric mean of positive a and b is √(ab).
If a, b, c are in GP (positive), then log a, log b, log c are in AP.
The product of the first n terms of a GP is (first × last)^n/2.

An AP grows by equal steps (linear); a GP with r > 1 grows faster and faster (exponential).

4. Means and the AM–GM–HM Inequality

For positive numbers a and b: AM = (a + b)/2, GM = √(ab), HM = 2ab/(a + b).

AM ≥ GM ≥ HM, with equality only when a = b; GM² = AM × HM

For n positive numbers: (a₁ + … + a_n)/n ≥ (a₁a₂…a_n)^1/n.

Using AM–GM for minimum values

If the product of positive terms is constant, their sum is least when the terms are equal. Example: for x > 0, x + 1/x ≥ 2√(x · 1/x) = 2, with the minimum 2 at x = 1. Likewise, if the sum is constant, the product is greatest when the terms are equal.

Inserting means

n arithmetic means between a and b: common difference d = (b − a)/(n + 1); their sum = n(a + b)/2.
n geometric means between a and b: common ratio r = (b/a)^1/(n+1); their product = (√(ab))ⁿ.

5. Special Series and the Method of Differences

Series	Sum
Σ k (k = 1 to n)	n(n + 1)/2
Σ k²	n(n + 1)(2n + 1)/6
Σ k³	[n(n + 1)/2]²
Σ k(k + 1)	n(n + 1)(n + 2)/3
Σ 1/(k(k + 1))	n/(n + 1)
Σ (2k − 1) (first n odd numbers)	n²

Method of differences (telescoping)

Write the general term as f(k) − f(k + 1). Then in the sum, the middle terms cancel and only f(1) − f(n + 1) remains. Example: 1/(k(k + 1)) = 1/k − 1/(k + 1).

Arithmetico-geometric series (AGP)

Terms like a, (a + d)r, (a + 2d)r², … To sum it, write S, multiply by r, subtract (S − rS) and sum the resulting GP. For |r| < 1:

S_∞ = a/(1 − r) + dr/(1 − r)²

6. Formula Master Table

Quantity	AP	GP
n-th term	a + (n − 1)d	arⁿ⁻¹
Sum of n terms	(n/2)[2a + (n − 1)d]	a(rⁿ − 1)/(r − 1)
Sum to infinity	—	a/(1 − r), \|r\| < 1
Condition for a, b, c	2b = a + c	b² = ac
Mean of a, b	(a + b)/2	√(ab)
n-th term from the end	l − (n − 1)d	l/rⁿ⁻¹

7. Mind Map

Sequences & Series

a + (n − 1)d
S_n = n/2(a + l)
a − d, a, a + d

arⁿ⁻¹
S_∞ = a/(1 − r)
a/r, a, ar

Means

AM ≥ GM ≥ HM
GM² = AM·HM
Min/max problems

Special series

Σn, Σn², Σn³
Telescoping
AGP

8. Tips and Tricks

Shortcuts

If the n-th term is given as a polynomial in n, write S_n by summing term by term with Σk, Σk², Σk³.
For "p-th term is q and q-th term is p" in an AP, the (p + q)-th term is 0.
For "S_p = S_q" in an AP (p ≠ q), S_p+q = 0.
In AM–GM problems with unequal coefficients, split terms into equal parts: e.g. 2x + 2x + 1/x² uses three terms whose product is constant.
For recurring decimals, write them as an infinite GP: 0.3̄ = 3/10 + 3/100 + … = 1/3.

Mistakes students make

Using S_∞ = a/(1 − r) when |r| ≥ 1 — the sum does not exist then.
Applying AM–GM to terms that can be negative. AM ≥ GM needs non-negative terms.
Claiming the minimum from AM–GM without checking that equality can actually occur.

9. Important One-liners

If a, b, c are in AP, then ka, kb, kc and a + k, b + k, c + k are also in AP.
If a, b, c are in GP, then 1/a, 1/b, 1/c are also in GP.
If a, b, c are in HP, then 1/a, 1/b, 1/c are in AP.
The sum of the first n odd natural numbers is n².
For positive x, x + 1/x ≥ 2 and equality holds only at x = 1.
The sum of an infinite GP with first term a and ratio r lies between a/2 and ∞ for 0 < r < 1.
In an AP, the sum of terms equidistant from the two ends is constant.
Σk³ = (Σk)²: the sum of cubes is the square of the sum.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Numerical

The sum of the first n terms of a sequence is S_n = 3n² + 5n. Find its 10th term.

Show solution

a_n = S_n − S_n−1 = 3n² + 5n − [3(n − 1)² + 5(n − 1)] = 6n + 2.

a₁₀ = 60 + 2 = 62. (Check: S₁₀ − S₉ = 350 − 288 = 62.)

Answer: 62

Q2Easy

The sum to infinity of a GP is 20 and its first term is 5. The common ratio is:

Show solution

a/(1 − r) = 20 with a = 5 ⇒ 1 − r = 1/4 ⇒ r = 3/4.

Answer: (B)

Q3Numerical

Find the minimum value of 4ˣ + 4¹⁻ˣ for real x.

Show solution

Both terms are positive. By AM ≥ GM: 4ˣ + 4¹⁻ˣ ≥ 2√(4ˣ · 4¹⁻ˣ) = 2√4 = 4.

Equality holds when 4ˣ = 4¹⁻ˣ, i.e. x = 1/2, so the minimum is attained.

Answer: 4

Q4Numerical

Find the value of 1·2 + 2·3 + 3·4 + … + 10·11.

Show solution

Σk(k + 1) = n(n + 1)(n + 2)/3. For n = 10: 10 × 11 × 12/3 = 440.

Answer: 440

Q5Medium

The sum 1/(1·2) + 1/(2·3) + … + 1/(99·100) equals:

1
99/100
100/101
1/100

Show solution

1/(k(k + 1)) = 1/k − 1/(k + 1). The sum telescopes: (1 − 1/2) + (1/2 − 1/3) + … + (1/99 − 1/100) = 1 − 1/100 = 99/100.

Answer: (B)

Q6Numerical

Find the sum to infinity of 1 + 2(1/2) + 3(1/2)² + 4(1/2)³ + …

Show solution

This is an AGP with a = 1, d = 1, r = 1/2. S_∞ = a/(1 − r) + dr/(1 − r)² = 1/(1/2) + (1/2)/(1/4) = 2 + 2 = 4.

(Also, 1 + 2x + 3x² + … = 1/(1 − x)², which gives 1/(1/4) = 4.)

Answer: 4

Q7Medium

Three numbers in AP have sum 15 and product 80. The largest of them is:

Show solution

Take them as a − d, a, a + d. Sum 3a = 15 ⇒ a = 5. Product 5(25 − d²) = 80 ⇒ d² = 9 ⇒ d = ±3.

The numbers are 2, 5, 8, so the largest is 8.

Answer: (A)

Q8Hard

If x and y are positive real numbers with x + y = 12, the maximum value of x²y is:

Show solution

Split x into two equal parts so that the three parts x/2, x/2 and y have a constant sum x/2 + x/2 + y = 12.

By AM–GM: 12/3 ≥ (x/2 · x/2 · y)^1/3 ⇒ x²y/4 ≤ 4³ = 64 ⇒ x²y ≤ 256.

Equality holds when x/2 = y = 4, i.e. x = 8, y = 4: then x²y = 64 × 4 = 256.

Answer: (A) 256

11. Quick Revision Summary

Remember before the exam

AP: a + (n − 1)d, S_n = n/2(a + l); GP: arⁿ⁻¹, S_∞ = a/(1 − r) for |r| < 1.
a_n = S_n − S_n−1 for any sequence.
AM ≥ GM ≥ HM for positive numbers; equality when all are equal.
Σk = n(n + 1)/2, Σk² = n(n + 1)(2n + 1)/6, Σk³ = [n(n + 1)/2]².
Telescoping: split into f(k) − f(k + 1); AGP: S − rS trick.

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