Sequences & Series

Sequences and series test AP/GP/HP mastery, special summations, and the powerful AM–GM–HM inequality. Advanced problems combine these — telescoping sums, arithmetico-geometric series, and inequality-based optimisation. Recognising the right summation trick is everything.

Key results

• AP: aₙ = a + (n−1)d; Sₙ = n/2(2a+(n−1)d) = n/2(first+last).
• GP: aₙ = ar^(n−1); Sₙ = a(rⁿ−1)/(r−1); infinite sum a/(1−r) for |r|<1.
• Means: AM ≥ GM ≥ HM (equal iff all terms equal); for two numbers GM² = AM·HM.
• Special sums: Σn = n(n+1)/2, Σn² = n(n+1)(2n+1)/6, Σn³ = [n(n+1)/2]².
• Techniques: telescoping (1/(n(n+1)) = 1/n − 1/(n+1)); arithmetico-geometric series via S − rS.

Series	Sum
Σn	n(n+1)/2
Σn²	n(n+1)(2n+1)/6
Σn³	[n(n+1)/2]²
GP (infinite)	a/(1−r),

Exam Tricks & Tips

• 🎯 AM ≥ GM ≥ HM turns many optimisation problems into one line — minima/maxima of expressions with a fixed sum or product follow immediately.
• 🎯 Telescoping: split the general term into a difference (partial fractions) so consecutive terms cancel — the go-to for 1/(n(n+1)) type sums.
• 🎯 For an arithmetico-geometric series, compute S − rS to collapse it to a GP plus a single term.
• 🎯 Between two numbers, GM is the geometric mean of AM and HM (GM² = AM·HM) — a quick relation for mean problems.
• 🎯 If terms are in AP and you need a GP/HP, transform (reciprocals give HP↔AP; logs give GP↔AP).
• ❌ Common mistake: using the infinite-GP formula a/(1−r) when |r| ≥ 1, where the series diverges and has no finite sum.

Expected exam pattern

1–2 questions: a special-summation/telescoping problem and an AM–GM inequality or AP/GP/HP relation (often integer or multiple-correct).

Quick recap

AP/GP/HP standard formulae; infinite GP a/(1−r) only for |r|<1. AM ≥ GM ≥ HM for optimisation, GM²=AM·HM for two numbers. Know Σn, Σn², Σn³. Telescope via partial fractions; handle AGP by S − rS.

Cover the answer, recall, then check. 12 cards on sequences and series for JEE Advanced.

Q1. Sum of the first n terms of an AP.
A1. Sₙ = n/2(2a + (n−1)d) = n/2(first + last).

Q2. Sum of an infinite GP and its condition.
A2. a/(1 − r), valid only for |r| < 1.

Q3. AM–GM–HM inequality.
A3. AM ≥ GM ≥ HM, with equality iff all terms are equal.

Q4. Relation between AM, GM, HM of two numbers.
A4. GM² = AM · HM.

Q5. Sum of the first n natural numbers squared.
A5. Σn² = n(n+1)(2n+1)/6.

Q6. Sum of the first n cubes.
A6. Σn³ = [n(n+1)/2]².

Q7. Telescoping form of 1/(n(n+1)).
A7. 1/n − 1/(n+1); consecutive terms cancel on summation.

Q8. How to sum an arithmetico-geometric series?
A8. Compute S − rS to reduce it to a GP plus one leftover term.

Q9. If a, b, c are in HP, what is in AP?
A9. Their reciprocals 1/a, 1/b, 1/c are in AP.

Q10. nth term of a GP.
A10. aₙ = ar^(n−1).

Q11. Condition for three numbers a, b, c to be in GP.
A11. b² = ac (b is the geometric mean).

Q12. Minimum of x + 1/x for x > 0 (via AM–GM).
A12. 2, attained at x = 1.

Sequences and series test pattern recognition and clever summation. JEE Advanced goes past AP/GP formulas into telescoping, arithmetico-geometric series, and the AM–GM–HM inequalities that pervade optimization problems across the syllabus.

Core concept: an AP has a constant difference, a GP a constant ratio; beyond these, most sums yield to telescoping (writing terms as differences) or standard power-sum formulas.

Core ideas

Beginner — AP and GP

AP: nth term a + (n−1)d; sum S_n = (n/2)(2a + (n−1)d) = (n/2)(first + last). GP: nth term ar^(n−1); sum S_n = a(rⁿ−1)/(r−1); infinite GP sums to a/(1−r) when |r| < 1. The three-term insertion tricks: for an AP take terms a−d, a, a+d; for a GP a/r, a, ar (symmetry simplifies the algebra).

Intermediate — means and special series

Means: between two positives, AM = (a+b)/2 ≥ GM = √(ab) ≥ HM = 2ab/(a+b), with equality iff a = b. Also GM² = AM·HM. Standard power sums: Σn = n(n+1)/2, Σn² = n(n+1)(2n+1)/6, Σn³ = (Σn)². These, with linearity, sum any polynomial series. Arithmetico-geometric series (each term = AP × GP, e.g. Σ n xⁿ) are summed by the "multiply by ratio and subtract" technique.

Advanced — telescoping and AM–GM optimization

Telescoping: if the general term t_n = f(n) − f(n+1), the sum collapses. Partial fractions create these: 1/(n(n+1)) = 1/n − 1/(n+1); 1/(n(n+1)(n+2)) splits similarly. Recognising a telescoping split is the single most powerful summation skill. AM–GM as an optimization tool: to minimise a sum of positive terms with fixed product (or maximise a product with fixed sum), AM ≥ GM gives the bound and the equality case gives the optimum — used constantly in "find the minimum value" problems, often faster than calculus. The general AM–GM for n terms: (a₁+…+aₙ)/n ≥ (a₁…aₙ)^(1/n).

Worked example

Find the sum Σ (k=1 to n) 1/(k(k+1)). Split by partial fractions: 1/(k(k+1)) = 1/k − 1/(k+1). The sum telescopes: (1/1 − 1/2) + (1/2 − 1/3) + … + (1/n − 1/(n+1)) = 1 − 1/(n+1) = n/(n+1). Every middle term cancels. As n → ∞ the sum → 1. Spotting the difference form turns a fearsome sum into a two-term answer — the essence of the telescoping method.

How JEE Advanced tests this

Summing series by telescoping (partial-fraction splits) and arithmetico-geometric technique; AP/GP with symmetric-term substitution; AM–GM/AM–GM–HM optimization (minimum/maximum values); mixed problems (terms forming AP whose reciprocals are in HP); and infinite-series convergence and sums.

Exam tricks & shortcuts

• Telescope: write t_n = f(n) − f(n+1) via partial fractions.
• AM ≥ GM ≥ HM (equality when all equal) — a fast optimization bound.
• Σn² = n(n+1)(2n+1)/6, Σn³ = (Σn)² — reduce polynomial sums.
• Arithmetico-geometric: multiply by r, subtract, sum the resulting GP.

Key formulas

• AP: aₙ = a + (n−1)d; Sₙ = (n/2)[2a + (n−1)d].
• GP: aₙ = ar^(n−1); Sₙ = a(rⁿ−1)/(r−1); S∞ = a/(1−r), |r| < 1.
• AM ≥ GM ≥ HM: (a+b)/2 ≥ √(ab) ≥ 2ab/(a+b); G² = AH.
• Special sums: Σn = n(n+1)/2; Σn² = n(n+1)(2n+1)/6; Σn³ = [n(n+1)/2]².

Use AGP/telescoping for mixed series; the AM–GM inequality bounds sums and products.

Worked Example

Problem: Find the sum of the first n terms of the series 1·2 + 2·3 + 3·4 + … , whose general term is k(k + 1).

Solution:
The general (k-th) term is Tₖ = k(k + 1) = k² + k. Sum from k = 1 to n using the standard formulas Σk² = n(n+1)(2n+1)/6 and Σk = n(n+1)/2:

Sₙ = Σ k² + Σ k
= n(n + 1)(2n + 1)/6 + n(n + 1)/2.

Factor out n(n + 1)/6:
= [n(n + 1)/6][(2n + 1) + 3]
= [n(n + 1)/6](2n + 4)
= [n(n + 1)/6]·2(n + 2)
= n(n + 1)(n + 2)/3.

Check for n = 3: 1·2 + 2·3 + 3·4 = 2 + 6 + 12 = 20; formula gives 3·4·5/3 = 20. ✓

Answer: Sₙ = n(n + 1)(n + 2)/3.