Geometric progressions

Fold a sheet of paper 0.1 mm thick in half: it becomes 0.2 mm thick. Fold again — 0.4 mm, then 0.8 mm, 1.6 mm, … Every fold doubles the thickness. After 10 folds it is only 102.4 mm, but after 42 folds — if that were possible — it would be 0.1 · 2⁴² mm ≈ 440,000 km. That is more than the average distance from the Earth to the Moon (about 384,000 km)! A real sheet cannot be folded that many times, but the arithmetic shows it: a quantity multiplied by the same number again and again grows incredibly fast.

In the lesson “Sequences and arithmetic progressions” we added the same number each time. Now we will multiply by the same number each time. In this lesson you will learn the formulas of geometric progressions, how to add infinitely many terms, how to turn repeating decimals into fractions, and why compound interest is a geometric progression.

Geometric progressions and the common ratio

Why b₁ ≠ 0 and r ≠ 0? Otherwise every term from the second on would be zero, and the ratio bₙ₊₁ ÷ bₙ could not be computed. The ratio decides how the progression behaves. For b₁ > 0: with r > 1 it increases (3, 6, 12, 24, …), with 0 < r < 1 it decreases (81, 27, 9, 3, …). With r < 0 the signs alternate (2, −6, 18, −54, …), and with r = 1 all terms are equal (7, 7, 7, …).

The nth term is found by counting steps, only now each step multiplies instead of adds: b₂ = b₁r, b₃ = b₁r², b₄ = b₁r³. From the first term to the nth there are n − 1 steps, so r appears as a factor n − 1 times.

When the first term and the ratio are not given, divide one known term by the other: from bₘ to bₙ there are n − m steps, and each step multiplies by r.

The characteristic property: the geometric mean

Take three consecutive terms bₙ₋₁, bₙ, bₙ₊₁. Neighbouring terms have the same ratio: bₙ ÷ bₙ₋₁ = bₙ₊₁ ÷ bₙ = r. Cross-multiplying this proportion gives bₙ² = bₙ₋₁ · bₙ₊₁. When the terms are positive, bₙ = √(bₙ₋₁ · bₙ₊₁): every term is the geometric mean of its neighbours, which is where the name comes from. The converse is also true: if this equality holds for every term of a sequence of non-zero numbers, the sequence is a geometric progression.

The sum of the first n terms

Gauss’s trick does not work here: in 1 + 2 + 4 + 8 the pairs 1 + 8 and 2 + 4 are not equal. There is another trick. Multiply Sₙ = b₁ + b₁r + b₁r² + … + b₁rⁿ⁻¹ by r: rSₙ = b₁r + b₁r² + … + b₁rⁿ. The two sums are almost the same: only b₁ is extra in the first and only b₁rⁿ in the second. Subtracting the first from the second cancels everything else: rSₙ − Sₙ = b₁rⁿ − b₁, that is, Sₙ(r − 1) = b₁(rⁿ − 1).

Infinite geometric series and repeating decimals

Suppose we walk 2 m like this: first 1 m, then half of the remaining distance (1/2 m), then half of what is left again (1/4 m), and so on. The distance walked is 1 + 1/2 + 1/4 + … Each step halves what is left, so the total gets as close to 2 as we like but never passes it. In formulas: if |r| < 1, then rⁿ gets closer and closer to 0 as n grows, so Sₙ = b₁(1 − rⁿ)/(1 − r) approaches b₁/(1 − r).

A repeating decimal is really the sum of an infinite geometric series. For example, 0.3̅ = 0.3 + 0.03 + 0.003 + …: here b₁ = 0.3 and r = 0.1, so 0.3̅ = 0.3 ÷ (1 − 0.1) = 0.3 ÷ 0.9 = 1/3. The equality 0.3̅6̅ = 4/11 from the lesson “Decimal fractions” is proved the same way.

Growth and decay problems

If a quantity changes by the same percentage in every period (a year, an hour, a bounce), its values form a geometric progression: an increase of p% multiplies by r = 1 + p/100 each time, a decrease of p% by r = 1 − p/100. The compound interest formula S = P · (1 + p/100)ⁿ from the lesson “Percentages” is exactly a term of a geometric progression: P is the first term, and the amount after n years is the (n + 1)th term.