Sequences and Series — JEE Main & Advanced Notes

AP, GP, HP, telescoping, sigma notation and recurrence patterns are high-return algebra topics.

Material signal: Mapped from sequence-series and mixed algebra revision files.

Priority: Scoring. Treat this as a foundation chapter inside the JEE Mathematics ladder.

• Identifying series type — the critical first step: The whole problem changes once you identify the series correctly. Compute aₙ₊₁−aₙ for a few terms — if constant, it's AP (common difference d). Compute aₙ₊₁/aₙ — if constant, it's GP (common ratio r). If neither, check second differences (constant → quadratic sequence), or try to split the term into AP·GP (arithmetic-geometric). Never guess the type; always verify with at least two consecutive terms.
• AP in depth — sum and structure: In an AP with first term a and common difference d, the nth term is aₙ = a+(n−1)d and the sum Sₙ = n/2·(first+last). The nth term from end is l−(n−1)d where l is the last term. Key trick: when three quantities are in AP, assign them as a−d, a, a+d — the sum is 3a (gives a immediately) and difference equations give d. Similarly, four terms in AP: a−3d, a−d, a+d, a+3d.
• GP in depth — ratio, power, convergence: aₙ = arⁿ⁻¹. The sum Sₙ = a(rⁿ−1)/(r−1) for r≠1, and = na for r=1. For an infinite GP, S∞ = a/(1−r) exists only when |r|<1. When r=−1, terms alternate and the series does not converge. For three numbers in GP, write a/r, a, ar — their product = a³ (gives a directly from the product condition), and the ratio condition gives r.
• AM–GM inequality — for optimisation: For positive numbers a₁,a₂,…,aₙ: AM=(a₁+⋯+aₙ)/n ≥ GM=(a₁⋯aₙ)^(1/n), with equality iff a₁=a₂=⋯=aₙ. In JEE: to find the minimum of an expression with a fixed sum, use GM≤AM: the product is maximised when all terms are equal. To find the minimum of a sum given a fixed product, use AM≥GM. The equality condition tells you the value of x at the optimum.
• Telescoping series — the collapse technique: A series Σaₖ telescopes when aₖ = f(k+1)−f(k) for some function f. Then Sₙ = f(n+1)−f(1) — all middle terms cancel in pairs. The key step is rewriting aₖ this way, often using partial fractions: 1/(k(k+1)) = 1/k−1/(k+1); 1/(k(k+2)) = ½(1/k−1/(k+2)). Recognise the pattern from the denominator structure.
• Method of differences for polynomial-type series: If aₙ is not AP or GP but the first-difference sequence bₙ = aₙ₊₁−aₙ is AP (or GP), write S = a₁+a₂+…+aₙ twice (shifted), subtract, and the result is a GP or AP sum. Repeat differencing if needed. This handles series like 1,4,9,16,… (sum = Σk²) or 1,3,7,13,21,… where the second differences are constant.
• Arithmetic-Geometric Progression (AGP): Terms are of the form (a+(k−1)d)·rᵏ⁻¹. To sum: write S = Σ(a+(k−1)d)rᵏ⁻¹. Multiply by r to get rS, then compute S−rS — the first part gives a GP (easier to sum) and the second gives a simpler AGP or GP. Repeat until you can evaluate directly. This S−rS trick is the standard and must be practised.
• Standard sigma sums and their proofs: Σk = n(n+1)/2 (pairing 1+n, 2+(n−1), etc.). Σk² = n(n+1)(2n+1)/6 (proved by telescoping k³−(k−1)³ and using Σk). Σk³ = [n(n+1)/2]² = (Σk)² — a remarkable identity. Memorise all three with the correct coefficients. For sums like Σ(2k²+3k+1), split and apply each formula term by term.

• AP: aₙ = a + (n−1)d; Sₙ = n/2·[2a + (n−1)d] = n/2·(first + last)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• GP: aₙ = ar^(n−1); Sₙ = a(rⁿ−1)/(r−1) for r≠1; S_∞ = a/(1−r) when |r|<1
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• HP: terms a₁,a₂,… are in HP iff 1/a₁, 1/a₂,… are in AP
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• AM ≥ GM ≥ HM for any set of positive numbers; equality iff all numbers are equal
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Σk = n(n+1)/2; Σk² = n(n+1)(2n+1)/6; Σk³ = [n(n+1)/2]²
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Telescoping: Σ[f(k+1)−f(k)] from k=1 to n = f(n+1)−f(1)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• AGP (arithmetic-geometric product): Tₙ = (a+(n−1)d)·rⁿ⁻¹; sum by S − rS method
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• AP three-number trick: write a−d, a, a+d (sum = 3a, product involves d)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• GP three-number trick: write a/r, a, ar (product = a³, immediately gives a)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.

• Applying the GP sum formula when r=1: Sₙ = a(rⁿ−1)/(r−1) is undefined at r=1. When r=1, every term equals a, so Sₙ = na. Always check whether r=1 before using the standard formula.
• Confusing Tₙ (nth term) with Sₙ (sum to n terms): Sₙ gives the running total; Tₙ = Sₙ−Sₙ₋₁ for n≥2 and T₁=S₁. Problems that give Sₙ and ask for Tₙ require subtraction — do not substitute n directly into the Sₙ formula and call it Tₙ.
• Applying AM≥GM to non-positive numbers: The AM–GM inequality requires all numbers to be positive (or non-negative for the weak form). If any number can be zero or negative, the inequality breaks down and the conclusion is invalid.
• Using S∞ formula without checking |r|<1: S∞ = a/(1−r) only converges for |r|<1. If r=2 or r=−1, writing S∞ produces a nonsensical answer. The condition |r|<1 must be verified explicitly.
• Wrong partial fractions in telescoping — re-derive every time: 1/(k(k+1)) = 1/k−1/(k+1) but 1/(k(k+2)) = ½·(1/k−1/(k+2)). Do not assume the same decomposition for different denominators; resolve partial fractions fresh each time.

For JEE Main, aim for fast recognition and clean substitution. Finish the first pass of Sequences and Series questions in 60–90 seconds each. Prioritise standard formulas, short sign/domain checks and option elimination only after the setup is correct.

For JEE Advanced, expect combined conditions, hidden domains, multi-correct traps and integer-style answers. Build the solution from definitions, not memorised tricks. When a parameter appears, solve the general case and then filter using restrictions.

• AP: constant first difference. Three unknowns → write a−d, a, a+d. Four unknowns → a−3d, a−d, a+d, a+3d.
• GP: constant ratio. Three unknowns → write a/r, a, ar. Product = a³ gives a instantly.
• AM≥GM: apply to positive numbers. Equality condition (all equal) gives the optimal point.
• Telescoping: rewrite Tₖ = f(k+1)−f(k) using partial fractions, then collapse to f(n+1)−f(1).
• Σk, Σk², Σk³: all three must be memorised with exact coefficients. Σk³ = (Σk)² is a quick sanity check.