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Math 20-1 · Sequences and series

Sequences and series

A sequence is a list; a series is what you get when you add the list up. Two kinds are worth knowing — the one that adds the same thing each step, and the one that multiplies — and each has a formula for the nth term and a formula for the sum.

The words, first

The idea: Most errors in this unit are a term formula used where a sum formula belonged, or an off-by-one in n − 1.

WordWhat it means
SequenceAn ordered list of numbers: 3, 7, 11, 15, …
SeriesThe sum of the terms of a sequence: 3 + 7 + 11 + 15.
Term / tnOne entry in the list. t₁ is the first, tₙ the general or nth term.
Common difference dWhat is added each step in an arithmetic sequence. Found by subtracting any term from the one after it.
Common ratio rWhat is multiplied each step in a geometric sequence. Found by dividing any term by the one before it.
ArithmeticConstant difference. Plotted, the terms sit on a straight line.
GeometricConstant ratio. Plotted, the terms curve like an exponential.
SnThe sum of the first n terms — a partial sum.
Convergent seriesAn infinite series whose partial sums approach a fixed number. Only geometric series with −1 < r < 1 do this.
Divergent seriesOne whose partial sums do not settle. Every arithmetic series with d ≠ 0 diverges.

Arithmetic sequences and series

The idea: Add d each step, so the nth term is the first term plus (n − 1) of them. There are n − 1 gaps between n terms, not n.

tₙ = t₁ + (n − 1)d  ·  Sₙ = n/2 [2t₁ + (n − 1)d]  ·  or Sₙ = n/2 (t₁ + tₙ) when you know the last term

Finding a term. t₁ = 5, d = 3: t₁₀ = 5 + 9(3) = 32. Nine steps, not ten.

Finding d from two terms. t₁ = 4 and t₈ = 32: 32 = 4 + 7d, so d = 4. Seven gaps between the first and eighth terms.

Summing. For 3, 7, 11, … the first 20 terms give S₂₀ = 10[6 + 19(4)] = 10(82) = 820. The second formula makes the reason clear: the average term is (3 + 79)/2 = 41, and 20 × 41 = 820.

Solving for n. If t₁ = 2, d = 3 and Sₙ = 155, then n(3n + 1)/2 = 155, so 3n² + n − 310 = 0 and n = 10. The negative root is discarded — a number of terms is a positive whole number.

Geometric sequences and series

The idea: Multiply by r each step. The exponent is n − 1 for the same reason as before, and the sum formula is a subtraction trick rather than something to memorise blindly.

tₙ = t₁ rn−1  ·  Sₙ = t₁(rⁿ − 1)/(r − 1), r ≠ 1

Finding a term. t₁ = 2, r = 3: t₅ = 2 × 3⁴ = 162. Using 3⁵ gives 486, which is the sixth term.

Summing. 2, 6, 18, … for six terms: S₆ = 2(3⁶ − 1)/(3 − 1) = 728. The sixth term alone is 486, so a total of 728 is the right size.

Geometric mean. If 5, x, 45 are consecutive, then x/5 = 45/x, so x² = 225 and x = ±15. Both work — one sequence has ratio 3 and the other −3 — so giving only 15 loses half the answer.

Infinite geometric series

The idea: Adding forever can still give a finite total, but only when each term is a genuine fraction of the one before it.

S = t₁/(1 − r), and only when −1 < r < 1

Worked example. 12 + 6 + 3 + 1.5 + … has r = ½, so S = 12/(1 − 0.5) = 24. The terms shrink fast enough that the total never passes 24.

Why the condition. At r = 1 the terms never shrink, so the sum grows without limit. At r = −1 they flip between two values forever and the partial sums never settle. Both endpoints are excluded, and anything outside diverges.

Repeating decimals. 0.3̄ = 0.3 + 0.03 + 0.003 + … is geometric with t₁ = 0.3 and r = 0.1, so S = 0.3/0.9 = ⅓. That is what a repeating decimal is.

What costs marks

The idea: Two of these are arithmetic slips; two are reading the question.

  • Using n instead of n − 1. The first term takes no steps to reach.
  • Mixing up a term and a sum. “The tenth term” and “the sum of ten terms” are different questions with different formulas.
  • Applying S when |r| ≥ 1. Check the ratio before using the formula; the answer is “it diverges”.
  • Giving only the positive geometric mean. x² = 225 has two solutions and both give a valid sequence.

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