Lesson 7: Sequences and Series

A sequence is an ordered list of numbers; a series is what you get when you add them. Two families dominate Algebra II — arithmetic sequences, which grow by constant differences, and geometric sequences, which grow by constant ratios — because they are discrete versions of the linear and exponential functions you already know. This lesson builds their term and sum formulas, sigma notation, and the surprising result that some infinite sums are finite.

Arithmetic sequences and series

An arithmetic sequence has a common difference d: a_n = a_1 + (n − 1)d. The sum of the first n terms pairs first with last: S_n = (n/2)(a_1 + a_n) = (n/2)(2a_1 + (n − 1)d).

Worked Example 1. For the sequence 7, 11, 15, 19, … (a_1 = 7, d = 4): the 20th term is a_20 = 7 + 19·4 = 83, and S_20 = (20/2)(7 + 83) = 10 · 90 = 900. Numerical sanity check by pairing: 7 + 83 = 11 + 79 = … = 90, and there are 10 such pairs — the formula is the pairing argument compressed. Graphically the terms lie on a line of slope d = 4: an arithmetic sequence is a linear function restricted to whole-number inputs.

Geometric sequences and series

A geometric sequence has a common ratio r: a_n = a_1·r^(n−1). The finite sum is S_n = a_1(1 − r^n)/(1 − r) for r ≠ 1.

Worked Example 2. For 3, 6, 12, 24, … (a_1 = 3, r = 2): a_8 = 3 · 2^7 = 384, and S_8 = 3(1 − 2^8)/(1 − 2) = 3(255) = 765. Check by adding: 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765 ✓. Graphically the terms lie on an exponential curve — Lesson 6 with the clock ticking in whole steps.

Sigma notation

The symbol Σ compresses a sum: the expression under it is the term formula, and the limits give the starting and ending index.

Worked Example 3. Evaluate Σ_{k=1}^{5} (2k + 1): substitute k = 1, 2, 3, 4, 5 to get 3 + 5 + 7 + 9 + 11 = 35. Notice this is also S_5 of an arithmetic sequence with a_1 = 3, d = 2: (5/2)(3 + 11) = 35 ✓ — two methods agreeing is the best error check in this lesson.

Infinite geometric series

If |r| < 1, the terms shrink toward zero fast enough that the infinite sum settles at S = a_1/(1 − r). If |r| ≥ 1, the sum does not settle (it diverges).

Worked Example 4. 12 + 6 + 3 + 1.5 + … has r = 1/2, so S = 12/(1 − 1/2) = 24. Partial sums — 12, 18, 21, 22.5, 23.25 — visibly home in on 24 without ever passing it.

Modelling: the bouncing ball and the savings plan

A ball is dropped from 10 m and rebounds to 60% of its previous height each bounce. It travels 10 m down, then up 6 and down 6, up 3.6 and down 3.6, and so on. Total distance:

10 + 2(6 + 3.6 + 2.16 + …) = 10 + 2 · 6/(1 − 0.6) = 10 + 2 · 15 = 40 metres

The ball bounces infinitely often yet travels a finite 40 m — the infinite geometric series is not a paradox, it is a measurement. (The first drop is counted once; every later height is travelled twice, up and down — modelling details like that factor of 2 are where the marks live.)

A saver deposits $200 at the end of each month into an account paying 0.4% monthly interest. The first deposit earns interest for 11 months, the last for none, so the value after 12 deposits is a geometric sum: 200(1.004^11 + 1.004^10 + … + 1) = 200(1.004^12 − 1)/(0.004) ≈ $2,453.51, against $2,400 paid in — $53.51 of interest, modest because most deposits were recent. This is the discrete engine inside every pension and mortgage calculation.

Common mistakes

• Off-by-one in the exponent or multiplier. From a_1 to a_20 there are 19 steps: a_20 = a_1 · r^19 in a geometric sequence, not r^20. Test formulas at n = 1: a wrong formula usually fails there immediately.
• Using the infinite-sum formula when |r| ≥ 1. 12 + 24 + 48 + … does not equal 12/(1 − 2) = −12; a sum of positives cannot be negative. Divergence first, formula second.
• Confusing d and r. A sequence can have differences 3, 3, 3 (arithmetic) or ratios 3, 3, 3 (geometric); "3, 6, 12" grows by ratio, "3, 6, 9" by difference. Diagnose from the data, not the vibe.
• Misreading sigma limits. Σ_{k=0}^{4} has five terms (k = 0, 1, 2, 3, 4). Count the index values.
• Sign errors with negative ratios. For r = −1/2, r^3 = −1/8 and terms alternate in sign; the formulas still work if you keep the signs.

Exam Experience

1. The 12th term of the arithmetic sequence 4, 9, 14, 19, … is

• (A) 54
• (B) 59
• (C) 64
• (D) 60

Answer: (B). d = 5, so a_12 = 4 + 11·5 = 59. (A) uses 10 gaps instead of 11 — the classic off-by-one; (C) uses 12.

2. The sum 5 + 10 + 20 + … + 640 (geometric, r = 2) equals

• (A) 1,275
• (B) 1,270
• (C) 645
• (D) 2,555

Answer: (A). 640 = 5·2^(n−1) gives 2^(n−1) = 128, so n = 8 terms; S_8 = 5(2^8 − 1) = 5 · 255 = 1,275.

3. The infinite series 8 + 8/3 + 8/9 + … converges to

• (A) 24
• (B) 16
• (C) 12
• (D) it diverges

Answer: (C). r = 1/3, |r| < 1, so S = 8/(1 − 1/3) = 8/(2/3) = 12.

Constructed response. A theatre has 20 rows of seats. The first row has 12 seats, and each row behind it has 3 more seats than the row in front. (a) How many seats are in the 20th row? (b) How many seats does the theatre hold in total? (c) The manager adds a 21st row continuing the pattern, then reports the new capacity as 810 + 75 = 885. Find and correct her error.

Scoring guide. (a) 1 point: arithmetic with a_1 = 12, d = 3; a_20 = 12 + 19·3 = 69 seats. (b) 2 points: 1 for choosing S_20 = (20/2)(a_1 + a_20); 1 for S_20 = 10(12 + 69) = 810 seats. (c) 2 points: 1 for computing a_21 = 12 + 20·3 = 72, not 75 (she added 3 for a 21st step from row 1, or extended from a miscounted row); 1 for the corrected capacity 810 + 72 = 882.

Going deeper: recursive definitions and disguised families

Sequences can be defined recursively — each term from its predecessors. The Fibonacci sequence a_1 = a_2 = 1, a_n = a_{n−1} + a_{n−2} gives 1, 1, 2, 3, 5, 8, 13, 21, …; no (n − 1)d or r^(n−1) formula applies directly, yet the sequence is perfectly determined. Recursion is how computer programs and biological populations often specify growth, and closed forms can hide surprisingly deep mathematics (Fibonacci's involves the golden ratio and √5).

A diagnostic challenge. The sequence 5, 9, 17, 33, 65, … is neither arithmetic (differences 4, 8, 16, 32) nor geometric in its terms (ratios 1.8, 1.89, 1.94, 1.97 — not constant, though converging). But the differences are geometric with ratio 2, and the pattern resolves if you subtract 1: 4, 8, 16, 32, 64 — pure powers of 2. So a_n = 2^(n+1) + 1: check n = 1: 4 + 1 = 5 ✓; n = 4: 32 + 1 = 33 ✓. Sequences of the form (geometric) + (constant) appear whenever a process doubles around a baseline — the same shape as the cooling law in Lesson 6 and the coffee model in Lesson 18. Learning to subtract the right constant to expose a clean pattern is a transferable modelling move, not a trick.

Series in medicine. A patient takes 50 mg of a drug daily; by the next dose, 30% of the previous amount remains in the body. Just after the nth dose the amount is 50 + 50(0.3) + 50(0.3)^2 + … + 50(0.3)^(n−1), a finite geometric series. As n grows it approaches the infinite sum 50/(1 − 0.3) ≈ 71.4 mg — the drug accumulates to a plateau about 43% above a single dose, not without bound. Pharmacologists call this the steady state; the infinite series computes it in one line, and the finiteness depends entirely on the ratio 0.3 being below 1. If 100% remained (ratio 1), the amount would grow linearly forever — the divergent case wearing a clinical face.

A pattern worth owning: the sum of the first n odd numbers

Compute small cases: 1 = 1; 1 + 3 = 4; 1 + 3 + 5 = 9; 1 + 3 + 5 + 7 = 16. The sums are perfect squares: S_n = n^2. The arithmetic-series formula confirms it instantly (a_1 = 1, d = 2: S_n = (n/2)(2 + (n − 1)·2) = (n/2)(2n) = n^2 ✓), and the picture is even better: each odd number is an L-shaped border that grows a square into the next square. So the sum of the first 50 odd numbers is 50^2 = 2,500 — a five-second answer where adding would take minutes. Series formulas are pattern compressors; this is the friendliest example of the genre and a reliable contest favourite.