Sequences and Series — AP, GP and HP

Progressions give NDA 3–5 questions and reward pure formula recall. AP and GP dominate; HP and the special sums appear occasionally. These are fast marks if the formulas are automatic.

Core ideas

Arithmetic Progression (AP): nth term aₙ = a + (n−1)d. Sum Sₙ = n/2 [2a + (n−1)d] = n/2 (first + last). Arithmetic mean of a and b is (a+b)/2.

Geometric Progression (GP): nth term aₙ = a·r^(n−1). Sum Sₙ = a(rⁿ−1)/(r−1) for r ≠ 1. Sum to infinity S∞ = a/(1−r), valid ONLY when |r| < 1. Geometric mean of a and b is √(ab).

Harmonic Progression (HP): reciprocals form an AP. Harmonic mean of a and b is 2ab/(a+b).

AM ≥ GM ≥ HM for positive numbers, with equality only when all terms are equal. Also GM² = AM × HM.

Special sums: Σn = n(n+1)/2; Σn² = n(n+1)(2n+1)/6; Σn³ = [n(n+1)/2]².

Exam Tricks & Tips

• 🎯 Choose symmetric terms: three in AP as a−d, a, a+d; three in GP as a/r, a, ar — this collapses the algebra.
• 🎯 In an AP, any term is the average of its neighbours: aₙ = (aₙ₋₁ + aₙ₊₁)/2.
• 🎯 S∞ = a/(1−r) applies ONLY for |r|<1; if |r|≥1 the infinite GP diverges — a favourite trap.
• 🎯 The sum of the first n odd numbers is n²; of the first n even numbers is n(n+1).
• 🎯 Σn³ = (Σn)² — the cube-sum is the square of the natural-number sum; memorise this shortcut.
• ❌ Common mistake: using Sₙ = a(rⁿ−1)/(r−1) when r = 1. If r = 1 the GP is constant and Sₙ = na.

Expected exam pattern

Expect: find a specified term or sum, insert means between two numbers, identify AP/GP/HP, or apply AM–GM. A recurring item gives two terms and asks for the common difference/ratio or a later term.

Quick recap

Memorise aₙ and Sₙ for AP and GP, the three means, AM ≥ GM ≥ HM with GM²=AM·HM, and Σn, Σn², Σn³. Watch the |r|<1 condition for infinite GP and the r=1 special case.

Cover, recall, check. 12 cards on progression formulas.

Q1. nth term of an AP?
A1. aₙ = a + (n−1)d.

Q2. Sum of n terms of an AP?
A2. Sₙ = n/2 [2a + (n−1)d] = n/2 (first + last term).

Q3. nth term of a GP?
A3. aₙ = a·r^(n−1).

Q4. Sum of n terms of a GP (r ≠ 1)?
A4. Sₙ = a(rⁿ − 1)/(r − 1).

Q5. Sum to infinity of a GP and its condition?
A5. S∞ = a/(1 − r), valid only for |r| < 1.

Q6. AM, GM, HM of two numbers a and b?
A6. AM = (a+b)/2, GM = √(ab), HM = 2ab/(a+b).

Q7. Relation among AM, GM, HM?
A7. AM ≥ GM ≥ HM, and GM² = AM × HM.

Q8. Σn (first n natural numbers)?
A8. n(n+1)/2.

Q9. Σn²?
A9. n(n+1)(2n+1)/6.

Q10. Σn³?
A10. [n(n+1)/2]² = (Σn)².

Q11. Sum of first n odd numbers?
A11. n².

Q12. Convenient way to take 3 terms in GP?
A12. a/r, a, ar (product = a³, simplifies fast).

Patterns of numbers — arithmetic, geometric, harmonic — and their sums form a dependable NDA scoring area with clean, memorable formulas.

Core idea / what this tests: Identifying the type of progression and applying the right n-th term and sum formula, plus the mean inequalities that link them.

Deep explanation

Beginner — Arithmetic Progression (AP)

Constant common difference d. n-th term aₙ = a + (n−1)d. Sum of n terms
Sₙ = n/2 · [2a + (n−1)d] = n/2 · (first + last).
Three terms in AP are conveniently taken as a−d, a, a+d.

Intermediate — Geometric Progression (GP)

Constant ratio r. n-th term aₙ = a·r^(n−1). Sum of n terms
Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1.
Infinite GP sums only when |r| < 1: S∞ = a/(1 − r).
Three terms in GP: a/r, a, ar.

Advanced — Harmonic Progression (HP) and means

Terms are in HP if their reciprocals are in AP. The n-th HP term = 1 / [a + (n−1)d] where a, d refer to the reciprocal AP.
For two positive numbers, the three means satisfy AM ≥ GM ≥ HM, with equality only when the numbers are equal, and the neat relation GM² = AM · HM.

• AM = (a+b)/2, GM = √(ab), HM = 2ab/(a+b).

Worked example

The third term of a GP is 12 and the sixth term is 96. Find the first term and ratio.
a·r² = 12 and a·r⁵ = 96. Divide: r³ = 96/12 = 8 ⇒ r = 2. Then a·4 = 12 ⇒ a = 3, r = 2. (Series: 3, 6, 12, 24, 48, 96 ✓)

NDA relevance

Common questions: find a specific term, sum a finite/infinite series, insert means between two numbers, or apply AM–GM to a "minimum value" problem. The AM ≥ GM inequality also appears in optimisation MCQs where calculus would be slower.

Exam tricks & shortcuts

• Symmetric picks (a−d, a, a+d) and (a/r, a, ar) kill the algebra when a product or sum of three terms is given.
• Sum of first n natural numbers = n(n+1)/2; squares = n(n+1)(2n+1)/6; cubes = [n(n+1)/2]².
• For "minimum of x + 1/x (x>0)", AM–GM gives 2 instantly.

Key formulas

• AP: aₙ = a + (n−1)d; Sₙ = n/2[2a + (n−1)d] = n/2(a + l).
• GP: aₙ = a rⁿ⁻¹; Sₙ = a(rⁿ−1)/(r−1); S∞ = a/(1−r), |r|<1.
• HP: reciprocals form an AP; nth term = 1/(a+(n−1)d).
• Means: AM = (a+b)/2, GM = √(ab), HM = 2ab/(a+b); AM ≥ GM ≥ HM.
• GM² = AM × HM.
• Σn = n(n+1)/2, Σn² = n(n+1)(2n+1)/6, Σn³ = [n(n+1)/2]².

Usage: identify AP (constant difference) vs GP (constant ratio) before choosing the sum formula.

Worked Example

Problem: In a geometric progression the 4th term is 24 and the 9th term is 768. Find the first term, the common ratio, and the sum of the first 8 terms.

Solution:
Let the first term be a and the common ratio be r. The nth term of a GP is aⁿ⁻¹ = a·r^(n−1).
4th term: a·r³ = 24 … (1)
9th term: a·r⁸ = 768 … (2)

Divide (2) by (1) to eliminate a:
r⁸ / r³ = 768 / 24 ⇒ r⁵ = 32 ⇒ r = 2.

Substitute r = 2 into (1): a·(2)³ = 24 ⇒ 8a = 24 ⇒ a = 3.

Sum of the first n terms of a GP (r ≠ 1) is Sₙ = a(rⁿ − 1)/(r − 1):
S₈ = 3(2⁸ − 1)/(2 − 1) = 3(256 − 1)/1 = 3 × 255 = 765.

Answer: a = 3, r = 2, and S₈ = 765.