🎯 Key Points
- AP: aₙ = a+(n-1)d, Sₙ = n/2[2a+(n-1)d]; GP: aₙ = a·rⁿ⁻¹, Sₙ = a(rⁿ-1)/(r-1), S∞ = a/(1-r) only when |r|<1
- AM-GM-HM: AM ≥ GM ≥ HM always, with equality ONLY when all terms are equal — a powerful inequality for optimization problems
- Key sums: Σn = n(n+1)/2, Σn² = n(n+1)(2n+1)/6, Σn³ = [n(n+1)/2]²
Sequences & Series
A sequence is an ordered list of numbers following a pattern. A series is the sum of a sequence.
Arithmetic Progression (AP)
Bar heights for AP 2, 5, 8, 11 with the common difference d=3 marked between consecutive terms.
- Common difference d between consecutive terms
- nth term: aₙ = a + (n-1)d
- Sum of n terms: Sₙ = n/2 [2a + (n-1)d] = n/2 (a + l)
- Arithmetic mean of a and b: (a+b)/2
Geometric Progression (GP)
Bar heights for GP 2, 4, 8, 16 doubling each time, common ratio r=2.
- Common ratio r between consecutive terms
- nth term: aₙ = a × rⁿ⁻¹
- Sum: Sₙ = a(rⁿ - 1)/(r - 1) for r ≠ 1
- Sum to infinity (|r| < 1): S∞ = a/(1-r)
- Geometric mean of a and b: √(ab)
Harmonic Progression (HP)
Reciprocals of an AP. Harmonic mean of a and b: 2ab/(a+b).
Important Sums
- Sum of first n naturals: n(n+1)/2
- Sum of first n squares: n(n+1)(2n+1)/6
- Sum of first n cubes: [n(n+1)/2]²
- Sum of first n odd numbers: n²
AM-GM-HM Inequality
For positive numbers: AM ≥ GM ≥ HM, with equality when all numbers are equal.
Properties of AP and GP
- If a constant is added to (or subtracted from) every term of an AP, the result is still an AP with the same common difference d.
- If every term of an AP is multiplied (or divided) by a non-zero constant k, the result is an AP with common difference kd.
- In a finite AP, terms equidistant from the two ends have a constant sum equal to (first + last): a₁ + aₙ = a₂ + aₙ₋₁ = ...
- Three numbers a, b, c are in AP if and only if 2b = a + c (b is the AM); they are in GP if and only if b² = ac (b is the GM).
- If every term of a GP is multiplied by a non-zero constant, or raised to the same power, the result is still a GP.
- Taking logarithms of the terms of a GP with positive terms produces an AP.
Inserting Arithmetic and Geometric Means
To insert n arithmetic means A₁, A₂, ..., Aₙ between two numbers a and b, build an AP of (n+2) terms with first term a and last term b. The common difference is d = (b − a)/(n+1), and Aₖ = a + k·d. The sum of these n AMs equals n(a+b)/2, i.e. n times the single AM of a and b.
To insert n geometric means G₁, ..., Gₙ between a and b, build a GP of (n+2) terms. The common ratio is r = (b/a)^(1/(n+1)), and Gₖ = a·rᵏ. The product of these n GMs equals (√(ab))ⁿ, the nth power of the single GM.
Relation Between AM and GM of Two Numbers
For two positive numbers a and b, let A = (a+b)/2 be their AM and G = √(ab) their GM. Then:
- A ≥ G always, with equality only when a = b, since A − G = (√a − √b)²/2 ≥ 0.
- a and b are the roots of x² − 2Ax + G² = 0, so a, b = A ± √(A² − G²) — this reconstructs the two numbers from their AM and GM.
- The AM, GM and HM of two numbers satisfy G² = A·H, so the GM is the geometric mean of the AM and the HM.
Convenient Selection of Terms
- 3 terms in AP: take a − d, a, a + d (their sum is 3a, eliminating d).
- 4 terms in AP: take a − 3d, a − d, a + d, a + 3d (common difference 2d).
- 3 terms in GP: take a/r, a, ar (their product is a³).
- 4 terms in GP: take a/r³, a/r, ar, ar³ (common ratio r²).
- This symmetric choice minimises the number of unknowns and is the fastest route for "sum and product" problems.
🚀 JEE Advanced Edge
Sum of an Arithmetico-Geometric Progression (AGP): A series like 1+3x+5x²+7x³+... (where the coefficient is an AP and the variable part is a GP) is solved by writing S, then xS shifted by one term, and subtracting (S - xS) to collapse most terms into a GP sum — this "multiply and subtract" trick is the standard JEE technique for AGP series that don't fit pure AP or GP formulas.
Telescoping sums via partial fractions: A series like Σ 1/[n(n+1)] simplifies because 1/[n(n+1)] = 1/n - 1/(n+1), so consecutive terms cancel in a chain, leaving only the first and last fragments — recognizing this telescoping pattern instantly solves sums that look intractable by direct AP/GP formulas.
Worked problem: Find the sum of the series 1/(1·2) + 1/(2·3) + 1/(3·4) + ... + 1/(99·100). Approach: Each term 1/[n(n+1)] = 1/n - 1/(n+1). The sum telescopes: (1/1-1/2)+(1/2-1/3)+...+(1/99-1/100) = 1/1 - 1/100 = 99/100.
Worked Example: Sum of a Product Series
Find the sum of the series 1·2 + 2·3 + 3·4 + … + n(n+1).
General term: k(k+1) = k² + k. Sum = Σk² + Σk = 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)(n+2)/3. Always split a product term into standard forms (Σk, Σk², Σk³) before summing.
Worked Example: Three Numbers in GP
Three numbers are in GP, their sum is 21 and their product is 216. Find the numbers.
Let the three terms be a/r, a, ar. Product = a³ = 216 → a = 6. Sum = 6/r + 6 + 6r = 21 → 6/r + 6r = 15 → 6r² − 15r + 6 = 0 → 2r² − 5r + 2 = 0 → (2r − 1)(r − 2) = 0 → r = 2 or r = ½.
Numbers: 3, 6, 12 (or 12, 6, 3). Taking a/r, a, ar with a = 6, r = 2 gives 3, 6, 12. Using a = 6, r = ½ gives the same set in reverse.