Lesson 13: Arithmetic and Geometric Progressions

Progressions are patterns with rules simple enough to compress into formulas. NCERT Class X studies the arithmetic progression fully — nth term and sum — while the geometric progression appears as its multiplicative twin, especially in exemplar and competency questions. The shared skill: translate "each step adds d" or "each step multiplies by r" into a formula, then solve for whichever of a, d (or r), n, aₙ, Sₙ is missing.

Arithmetic progression

An AP has first term a and common difference d. The nth term is aₙ = a + (n − 1)d — note the (n − 1): the first term has had *zero* differences added. The sum of the first n terms is Sₙ = (n/2)[2a + (n − 1)d], equivalently Sₙ = (n/2)(a + l) where l is the last term.

Worked example 1 (basic). For the AP 5, 8, 11, …, find the 20th term and the sum of the first 20 terms. a₂₀ = 5 + 19 × 3 = 62. S₂₀ = (20/2)[2(5) + 19(3)] = 10 × 67 = 670. Cross-check with the other form: (20/2)(5 + 62) = 10 × 67 ✓ — use one formula to check the other on exams.

Worked example 2 (intermediate — which term?). Which term of the AP 3, 8, 13, 18, … is 78? 3 + (n − 1)(5) = 78 → (n − 1) = 15 → n = 16. Every "which term" problem is this linear equation; if n comes out non-integral or negative, the number simply is not a term of the AP — a conclusion exemplar questions sometimes want stated explicitly.

Geometric progression

A GP has ratio r: aₙ = arⁿ⁻¹ and Sₙ = a(rⁿ − 1)/(r − 1) for r > 1. For 3, 6, 12, …: a₈ = 3 × 2⁷ = 384; S₅ = 3(2⁵ − 1)/(2 − 1) = 93. GPs model anything that scales: population, compound interest (the most important AP/GP boundary case — interest *compounds*, so money follows a GP, not an AP), bacterial doubling, and depreciation.

Competency application: the savings plan

A student saves ₹100 in January and increases her saving by ₹50 every month: 100, 150, 200, … How much does she save in the whole year? This is an AP with a = 100, d = 50, n = 12: S₁₂ = (12/2)[200 + 11 × 50] = 6 × 750 = ₹4500. Contrast for the case set's follow-up: her brother invests ₹100 at 10% annual compound interest — a GP with r = 1.1, worth 100 × 1.1¹² ≈ ₹313.84 after 12 years, against a linear ₹100 + 12 × ₹10 = ₹220. The follow-up question "which grows faster and why" tests the AP/GP distinction, not arithmetic.

A sum to know. Sum of the first 15 multiples of 8: a = 8, d = 8, n = 15: S = (15/2)[16 + 14 × 8] = (15/2)(128) = 960.

Common mistakes

• Writing aₙ = a + nd. The first term uses zero differences. Off-by-one errors here contaminate every subsequent line.
• Using the AP sum formula on a GP scenario — compound interest, doubling bacteria, and bouncing balls (each bounce a fixed *fraction* of the previous height) are all GPs.
• In "which term" problems, accepting n = 16.5 as an answer. A term number must be a positive integer; a fraction means the value does not occur.
• Forgetting there are two Sₙ forms. If the last term is given, Sₙ = (n/2)(a + l) is faster and less error-prone.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. The common difference of the AP 4, 9, 14, 19, … is: (a) 4 (b) 5 (c) 9 (d) −5. Answer: (b) 5. Each term exceeds the previous by 5.

Q2. The 10th term of the AP 2, 7, 12, … is: (a) 47 (b) 52 (c) 45 (d) 50. Answer: (a) 47. 2 + 9 × 5 = 47.

Section B — Short answer (3 marks) Q3. Find the sum of the first 15 multiples of 8. Marking scheme: 1 mark for identifying a = 8, d = 8, n = 15; 1 mark for Sₙ = (n/2)[2a + (n − 1)d] = (15/2)[16 + 112]; 1 mark for the value 960.

Section C — Case-based long answer (5 marks) Q4. A school plants roses in rows: the first row has 23 plants, the second 21, the third 19, and so on (2 fewer each row), for 10 rows. Separately, a gardener trialling a new fertiliser finds the number of flowers on a bush triples every week from an initial 5 flowers, for 4 weeks of observations (weeks 0 to 3). (i) Which sequence do the rose rows follow, and what are a and d? (1 mark) (ii) How many rose plants are there in all 10 rows? (2 marks) (iii) How many flowers are there in week 3, and what is the total counted over weeks 0–3? (2 marks) Marking scheme: (i) AP with a = 23, d = −2 (1 mark). (ii) S₁₀ = (10/2)[46 + 9(−2)] = 5 × 28 = 140 plants (1 mark for the set-up, 1 for the value). (iii) Week 3 = 5 × 3³ = 135 flowers (1 mark); total = 5 + 15 + 45 + 135 = 200, or by formula 5(3⁴ − 1)/(3 − 1) = 5 × 40 = 200 (1 mark).

Going deeper: sum of the first n odd numbers

The AP 1, 3, 5, …, (2n − 1) has Sₙ = (n/2)[2 + (n − 1)2] = (n/2)(2n) = n². So the first 10 odd numbers sum to 100, the first 25 to 625 — and a picture explains it: each odd number adds an L-shaped border that grows a square into the next square. Similarly the first n even numbers sum to n(n + 1). Results like these are one-mark gifts when recognised and five-minute computations when not.

Going deeper: common terms of two APs

The APs 2, 5, 8, 11, … (d = 3) and 3, 7, 11, 15, … (d = 4) share the term 11. Their common terms form a new AP whose difference is lcm(3, 4) = 12: 11, 23, 35, … with nth common term 11 + (n − 1) × 12. The method generalises: find the *first* common term by inspection or simultaneous congruences, then step by the LCM of the differences. Exemplar sets use this to ask for, e.g., the 10th common term: 11 + 9 × 12 = 119 (check: 119 = 2 + 39 × 3 ✓ and 119 = 3 + 29 × 4 ✓).

Going deeper: geometric means and the infinite GP

To insert two geometric means between 2 and 54, find r with 2r³ = 54: r³ = 27, r = 3, giving 2, 6, 18, 54. For |r| < 1 the GP has an infinite sum S∞ = a/(1 − r): 1 + 1/2 + 1/4 + … = 1/(1 − 1/2) = 2. The showpiece application: a ball dropped from 10 m rebounds to 3/4 of its height each time. Total distance travelled = 10 (down) + 2 × [7.5 + 7.5(3/4) + …] (up and down after each bounce) = 10 + 2 × 7.5/(1 − 3/4) = 10 + 2 × 30 = 70 m. One formula converts an infinite journey into arithmetic.

One more inequality: AM ≥ GM

For two positive numbers, the arithmetic mean dominates the geometric mean: (a + b)/2 ≥ √(ab), with equality only when a = b. For 4 and 16: AM = 10, GM = 8 ✓. The proof is one line: (a + b)/2 − √(ab) = (√a − √b)²/2 ≥ 0, since a square is never negative. This humble inequality powers optimisation results across mathematics (it is the silent engine behind Lesson 19's equal-split principle: for fixed sum, the product ab is maximised when a = b, because √(ab) ≤ the fixed AM). In a GP, the same geometry appears as the middle-term property: each term is the geometric mean of its neighbours, e.g. 8² = 4 × 16 in 4, 8, 16, … A final pattern worth filing: the three-term template. If three numbers are in AP, write them a − d, a, a + d (the sum is instantly 3a); if in GP, write a/r, a, ar (the product is instantly a³). Problems that *give* the sum of three AP numbers or the product of three GP numbers are engineered for these templates — choosing symmetric unknowns converts two equations in two unknowns into one equation in one unknown, and is a recognised exemplar-level move.

Final drills

Drill 1 — a shrinking AP. For the AP 100, 96, 92, …, find the 25th term: a₂₅ = 100 + 24 × (−4) = 4. Negative differences are routine; the formula does not care. Similarly, the sum of an AP given first and last terms only: if a = 5, l = 95 and n = 30, then S₃₀ = (30/2)(5 + 95) = 1500 — no need for d at all.

Drill 2 — a fractional GP. The GP 81, 27, 9, … has r = 1/3. Its 6th term is 81 × (1/3)⁵ = 81/243 = 1/3. Sum of the first 4 terms: 81 + 27 + 9 + 3 = 120 — verify by formula: 81 × (1 − (1/3)⁴)/(1 − 1/3) = 81 × (80/81) × (3/2) = 120 ✓. And a "which term" for GP: in 2, 8, 32, …, which term is 512? 2 × 4ⁿ⁻¹ = 512 gives 4ⁿ⁻¹ = 256 = 4⁴, so n = 5. GP term-hunting is exponent-matching; AP term-hunting was linear solving — the two chapters rhyme.