Lesson 19: Sequences: Arithmetic and Geometric

A sequence is a function whose domain is the positive integers: a list a₁, a₂, a₃, … where position matters. Two families dominate. An arithmetic sequence adds the same common difference d at each step; a geometric sequence multiplies by the same common ratio r. They are discrete versions of the linear and exponential functions — a sequence is just a function you meet one term at a time.

Arithmetic sequences

Terms like 7, 11, 15, 19, … have d = 4. To reach the nth term from the first you take n − 1 steps (there are n − 1 gaps between n terms — the off-by-one fact that runs this whole lesson), so: aₙ = a₁ + (n − 1)d. For our sequence, a₂₀ = 7 + 19 × 4 = 83. Running the formula backwards finds positions: which term equals 67? 7 + 4(n − 1) = 67 → 4(n − 1) = 60 → n − 1 = 15 → n = 16 (check: a₁₆ = 7 + 60 = 67 ✓). Sum of the first n terms. Pair first with last, second with second-last — every pair sums to a₁ + aₙ (Gauss's schoolroom trick): Sₙ = (n/2)(a₁ + aₙ). Here S₂₀ = (20/2)(7 + 83) = 10 × 90 = 900. The pairing logic: 20 terms make 10 pairs, each totalling 90.

Geometric sequences

Terms like 3, 6, 12, 24, … have r = 2. Each step multiplies, so the nth term carries n − 1 factors of r: aₙ = a₁ · rⁿ⁻¹. Here a₈ = 3 × 2⁷ = 384. Ratios need not be whole: 81, 54, 36, … has r = 54/81 = 2/3; the next term is 36 × 2/3 = 24, and a₅ = 81 × (2/3)⁴ = 81 × (16/81) = 16. A ratio between 0 and 1 creates decay — the discrete twin of Lesson 12's exponential decay, and indeed aₙ = a₁rⁿ⁻¹ *is* y = a·bˣ sampled at integer points. Sequences can also be defined recursively (each term from the previous: aₙ = aₙ₋₁ + 4, a₁ = 7) rather than explicitly (aₙ in terms of n alone). Recursive definitions are natural for processes; explicit formulas are what you want for jumping straight to the 100th term.

Modelling application 1: a savings plan

You open an account with $200 and deposit $35 at the end of every week. The balance after n weeks is Bₙ = 200 + 35n — arithmetic growth built on a head start. After 12 weeks: 200 + 420 = $620. When does the balance pass $1,000? 200 + 35n > 1000 → 35n > 800 → n > 22.86, so during week 23 (check the neighbours: week 22 gives $970, week 23 gives $1,005 ✓). Note the model's index care: "after n weeks" — the deposits, not the calendar weeks, are the steps of the sequence.

Modelling application 2: the bouncing ball

A ball dropped from 10 m rebounds to 80% of its previous height each bounce. The peak heights form a geometric sequence with r = 0.8: 10, 8, 6.4, 5.12, … so after the third bounce the peak is 5.12 m (10 × 0.8³). The heights shrink toward zero but — mathematically — never reach it: aₙ = 10(0.8)ⁿ stays positive for every n. Real balls stop because friction and deformation eat more than the model assumes; the geometric model is excellent for the first several bounces and a polite fiction thereafter. Naming where a model stops being true is part of using it well.

Common mistakes

• The off-by-one in (n − 1). a₂₀ uses 19 gaps, not 20: a₁ has taken zero steps. Students who write a₁ + 20d are computing a₂₁. Sanity-test formulas at n = 1: a₁ + (1 − 1)d = a₁ ✓; a₁ · r⁰ = a₁ ✓.
• Using the difference formula on a geometric sequence (or the ratio on an arithmetic one). Diagnose the family first — subtract consecutive terms: constant gaps → arithmetic. Divide consecutive terms: constant ratio → geometric. Some sequences (1, 2, 4, 7, 11, … — gaps 1, 2, 3, 4) are neither.
• Sign errors with negative differences. 2, −1, −4, … has d = −3, so a₁₀ = 2 + 9(−3) = −25. Writing d = 3 because "the terms change by 3" loses the direction of travel.
• In geometric sums and terms, rⁿ vs rⁿ⁻¹. a₅ for 3, 6, 12, … is 3 × 2⁴ = 48, not 3 × 2⁵ = 96 (that's a₆). The exponent counts *multiplications from the first term* — again n − 1.

Exam Experience

Question 1. What is the 15th term of the arithmetic sequence 2, 5, 8, …? A. 47 B. 44 C. 32 D. 45 Answer: B. d = 3; a₁₅ = 2 + 14 × 3 = 44. Option A uses 15 gaps instead of 14 — the off-by-one trap in its natural habitat.

Question 2. A geometric sequence has a₁ = 5 and r = 3. Its 4th term is: A. 20 B. 45 C. 135 D. 405 Answer: C. a₄ = 5 × 3³ = 5 × 27 = 135. Option B multiplies by 3 only twice... (5 × 9 = 45, the 3rd term); D applies one factor too many.

Question 3. The sequence 81, 54, 36, 24, … is: A. Arithmetic with d = −27 B. Geometric with r = 2/3 C. Geometric with r = 3/2 D. Neither arithmetic nor geometric Answer: B. Differences are −27, −18, −12 (not constant), but ratios are 54/81 = 36/54 = 24/36 = 2/3 throughout. Testing *both* operations is the diagnosis.

Question 4 (constructed response). A training plan starts with a 2-mile run in week 1 and increases the long run by 0.5 miles each week. (a) Write an explicit formula for the long-run distance aₙ in week n. (b) Find the long-run distance in week 16. (c) Find the total mileage of the long runs over the 16 weeks. (d) In which week does the long run first reach 8 miles?

Scoring guide (4 points).

• 1 point: arithmetic with a₁ = 2, d = 0.5: aₙ = 2 + 0.5(n − 1).
• 1 point: a₁₆ = 2 + 0.5(15) = 9.5 miles.
• 1 point: S₁₆ = (16/2)(2 + 9.5) = 8 × 11.5 = 92 miles.
• 1 point: 2 + 0.5(n − 1) = 8 → 0.5(n − 1) = 6 → n − 1 = 12 → week 13.

Deeper still: summing a geometric sequence, and pay over a year

Geometric sums have a formula worth the small derivation: for Sₙ = a₁ + a₁r + … + a₁rⁿ⁻¹, multiplying by r and subtracting telescopes everything away except the ends, leaving Sₙ = a₁(rⁿ − 1)/(r − 1) (for r ≠ 1). For 3 + 6 + 12 + … + 384 (8 terms, r = 2): S₈ = 3(2⁸ − 1)/(2 − 1) = 3 × 255 = 765 — confirmed by direct addition: 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765 ✓. The famous chessboard legend trades on this formula: grains doubling per square give 2⁶⁴ − 1 ≈ 1.8 × 10¹⁹ grains in total — more than the world's wheat harvest for millennia — from nothing but r = 2 and 64 terms. Exponentials punish intuition; formulas keep you honest. Arithmetic sums close the course's loop back to money. A job pays $500 in week 1, rising by $20 each week for 52 weeks. The final week's pay is a₅₂ = 500 + 51 × 20 = $1,520, and the year's total is S₅₂ = (52/2)(500 + 1520) = 26 × 2020 = $52,520. Notice the elegance: the average week's pay is (500 + 1520)/2 = $1,010 — the mean of first and last — times 52 weeks. That is exactly Gauss's pairing wearing a payslip: for arithmetic growth, the overall average is always the average of the endpoints.

One more idea: Fibonacci, triangular numbers, and second differences

Not every pattern is arithmetic or geometric. The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, … is defined *recursively*: each term is the sum of the previous two (aₙ = aₙ₋₁ + aₙ₋₂). Its ratios — 8/5 = 1.6, 13/8 = 1.625, 21/13 ≈ 1.615 — hover ever closer to the golden ratio ≈ 1.618, which is why Fibonacci patterns surface in sunflower spirals and pinecones: nature's growth processes are recursive, each new structure built from the recent ones. The triangular numbers 1, 3, 6, 10, 15, … (dot patterns forming triangles) have differences 2, 3, 4, 5 — not constant, so not arithmetic — but the *second* differences are constant (all 1), the fingerprint of a quadratic rule: aₙ = n(n + 1)/2 (check: n = 4 gives 4 × 5/2 = 10 ✓). And here the course closes a loop: n(n + 1)/2 is also the sum 1 + 2 + … + n — Gauss's pairing from the arithmetic-sum formula, Sₙ = (n/2)(1 + n), in disguise. Sequences, series, quadratics: the families were never separate countries, just neighbouring towns with good roads between them.

A last diagnostic drill assembles the whole family test. The sequence 5, 10, 20, 35 has differences 5, 10, 15 — not constant — and ratios 2, 2, 1.75 — not constant either — so it is neither arithmetic nor geometric. But the differences grow arithmetically (second differences all 5), the quadratic fingerprint, and indeed aₙ = (5/2)n(n − 1) + 5 reproduces it: n = 1 → 5, n = 2 → 10, n = 3 → 20, n = 4 → 35. Keep the flowchart by the door: constant differences → arithmetic; constant ratios → geometric; constant *second* differences → quadratic. Nearly every pattern question in algebra is one of these three in costume.