Recurrence Relations in JEE Main Mathematics

Recurrence relations define sequences where each term is expressed as a function of previous terms. JEE Main tests them disguised as "sequence of function values" or "recursive formula" problems — students who recognise them solve in 90 seconds; others try to enumerate terms until they spot a pattern, taking five minutes and making errors. This guide gives you the tools to solve them systematically.

First-Order Linear Recurrences

Form: a_{n+1} = r × a_n + d (arithmetic-geometric type). Solution: if r = 1, a_n = a_1 + (n−1)d (arithmetic sequence). If r ≠ 1: find the fixed point L = d/(1−r), then a_n − L = r^(n-1) × (a_1 − L). General solution: a_n = L + r^(n-1)(a_1 − L) where L = d/(1−r). Example: a_{n+1} = 3a_n + 2, a_1 = 1. Fixed point: L = 2/(1−3) = −1. a_n = −1 + 3^(n-1)(1−(−1)) = −1 + 2×3^(n-1). Check: a_2 = −1 + 2×3 = 5. Verify: 3×1+2 = 5 ✓. JEE question: given the recurrence and a_1, find a_n or a specific term.

When d = 0 (purely geometric): a_n = a_1 × r^(n-1). When r = 1 (purely arithmetic): standard AP formula. When both r ≠ 1 and d ≠ 0: use the fixed-point method above. Take a free sequences mock. See our sequences and series guide.

Second-Order Linear Recurrences: Characteristic Equation

Form: a_{n+2} = p × a_{n+1} + q × a_n. Characteristic equation: r² = pr + q → r² − pr − q = 0. If roots r₁, r₂ are distinct: a_n = A r₁^n + B r₂^n. If roots are equal (r₁ = r₂ = r): a_n = (A + Bn)r^n. Fibonacci sequence: a_{n+2} = a_{n+1} + a_n → characteristic equation r² − r − 1 = 0 → roots r = (1±√5)/2. General term: a_n = [φⁿ − ψⁿ]/√5 where φ = (1+√5)/2 (golden ratio) and ψ = (1−√5)/2. JEE does not ask you to derive Fibonacci but uses Fibonacci-like patterns in sequence questions.

Recurrences in Counting and Probability

Recurrence relations naturally arise in combinatorics and probability. Example: Number of ways to tile a 1×n board with 1×1 and 1×2 tiles. Let a_n = number of ways. For the last tile: if it's 1×1, the remaining board has a_{n-1} ways; if it's 1×2, remaining board has a_{n-2} ways. So a_n = a_{n-1} + a_{n-2} with a_1 = 1, a_2 = 2. This is Fibonacci starting from a_1 = 1, a_2 = 2. JEE uses such counting recurrences in combinatorics questions.

Probability recurrences: P_n = probability of event at step n. Example: a frog jumps 1 or 2 steps forward. Probability of reaching step n = P_{n-1} × (prob of 1-step jump) + P_{n-2} × (prob of 2-step jump). If both jumps are equally likely (1/2 each): P_n = (1/2)P_{n-1} + (1/2)P_{n-2}. Characteristic equation: 2r² − r − 1 = 0 → (2r+1)(r−1) = 0 → r = 1 or r = −1/2. General solution P_n = A + B(−1/2)^n. As n → ∞, P_n → A (convergence condition determines A and B from initial conditions). For AP/GP series summation see our advanced sequences guide.