Drop a bouncy ball from 2 metres and watch what it does. It bounces back up — but not all the way. A decent rubber ball keeps about 60\% of its height on each bounce: 2 m, then 1.2 m, then 0.72 m, then 0.432 m… Each height is the one before it multiplied by 0.6.

That "multiply by the same number each step" pattern is everywhere once you start looking. A virus that infects three new people per case triples each week: 1,\ 3,\ 9,\ 27,\ 81,\ \dots A savings account paying 5\% interest turns £1000 into £1050, then £1102.50 — each balance is the last one times 1.05. None of these grows (or shrinks) by adding a fixed amount; they all multiply by a fixed amount.

And here is the shock this page is building towards: the ball bounces infinitely many times (each bounce is smaller, but there's always another one) — yet the total distance it travels is a perfectly ordinary, finite number. You will be able to compute it, exactly, before the end of the page.

The curve never quite reaches zero — mathematically the ball bounces forever — but the heights die away so fast that, as we'll see, all those infinitely many bounces add up to something finite.

The definition: a fixed ratio, not a fixed difference

In an arithmetic sequence you add the same amount each step. A geometric sequence does the same trick with multiplication: you multiply by the same number — the common ratio r — to get from one term to the next.

Starting from a first term a, the terms are

Each term carries one more factor of r than the last, so the nth term is

The power is n-1, not n, because the first term a has been multiplied by r zero times — the index laws let us write that build-up of factors as a single power.

Two quick tests you can run on any sequence. To check it's geometric, divide each term by the one before: if \dfrac{a_2}{a_1} = \dfrac{a_3}{a_2} = \dfrac{a_4}{a_3} = \dots you have a constant ratio. To find r, one such division is all you need: for 4,\ 12,\ 36,\ 108,\ \dots we get r = 12 \div 4 = 3 (and 36 \div 12 = 3 confirms it).

Growth and decay: what r controls

Take a = 3 and r = 2:

The 4th term is a\,r^{\,3} = 3 \times 2^3 = 24 — no need to step through all of them. That's the whole point of the formula: it teleports you straight to any term. The 20th term is 3 \times 2^{19} = 1\,572\,864, found in one line rather than nineteen doublings.

The size of r decides the sequence's whole character. With r > 1 the terms grow — and not politely: multiplying compounds, so a geometric sequence with r = 2 soon dwarfs any arithmetic sequence, however generous its common difference. With 0 < r < 1 each term is a fraction of the last, so they shrink toward zero — the bouncing ball's fading heights, or a medicine dose decaying in the bloodstream. Watch both behaviours side by side:

Both curves start at the same value, a = 1. The r = 2 curve doubles at every step and rockets off the top of the chart by n = 5; the r = \tfrac12 curve halves at every step and hugs the axis, getting ever closer to zero without ever touching it. Same rule, opposite destinies — all decided by whether |r| is bigger or smaller than 1.

Worked example: which term first beats 1000?

Here's the classic exam question. In the sequence 3,\ 6,\ 12,\ 24,\ \dots (so a = 3, r = 2), which term is the first to exceed 1000?

We need the smallest n with

Method 1 — trial. Run up the powers of two: 2^8 = 256 (too small), 2^9 = 512 (big enough). So n - 1 = 9, giving n = 10.

Method 2 — logarithms. Taking logs of both sides of 2^{\,n-1} > 333.3:

so n - 1 = 9 is the first whole number that works, and again n = 10. Check it: a_{10} = 3 \times 2^9 = 1536 > 1000, while a_9 = 3 \times 2^8 = 768 falls short. The 10th term is the first past the post. Trial is fine when r is a friendly whole number; logs are the general-purpose tool when it isn't (try finding when 1.05^{\,n} first doubles your money by trial and you'll wish you'd used logs).

Adding the terms: a geometric series

A series is the sum of the terms. Adding the first n terms of a geometric sequence gives

There's a closed form for this, and the way it's derived is one of the loveliest tricks in algebra — worth seeing in full. Multiply the whole sum by r: every term slides one place to the right,

Now subtract. The two rows share almost every term — ar, ar^2, \dots, ar^{\,n-1} all appear in both and cancel in one satisfying telescope — leaving only the first term of the top row and the last term of the bottom:

Factorise both sides — S_n(1 - r) = a(1 - r^{\,n}) — and divide:

A sum of a thousand terms, captured by three symbols. (When r = 1 every term is just a, so S_n = na and no formula is needed.)

Check it on something small. For 3, 6, 12, 24 with a = 3, r = 2, n = 4: S_4 = \dfrac{3(1 - 2^4)}{1 - 2} = \dfrac{3 \times (-15)}{-1} = 45, and indeed 3 + 6 + 12 + 24 = 45.

Now something big. Sum the first ten terms of 5 + 10 + 20 + 40 + \dots Here a = 5, r = 2, n = 10:

Adding ten terms by hand invites ten chances to slip; the formula does it in one move. A tidy habit for r > 1: both brackets come out negative, and the minuses cancel — many people flip the formula to S_n = \frac{a(r^{\,n} - 1)}{r - 1} for growing series so everything stays positive. It's the same formula with top and bottom each multiplied by -1.

The sum to infinity

Something special happens when |r| < 1. Each term is smaller than the last, so the running total piles up less and less — and the partial sums S_n creep toward a fixed value rather than running away. Look at the formula we just derived: the only part that depends on n is r^{\,n}, and when |r| < 1 that power crumbles away — (\tfrac12)^{10} \approx 0.001, (\tfrac12)^{20} \approx 0.000001 — so r^{\,n} \to 0 as n \to \infty, leaving the sum to infinity:

The classic case is a = 1, r = \tfrac{1}{2}:

Watch the partial sums below: each new term adds half of the gap that remains, so the total edges ever closer to 2 without ever passing it. (If |r| \ge 1 the terms do not shrink, the total never settles, and there is no sum to infinity.)

This is a genuinely deep idea: infinitely many positive numbers can add to a finite total, provided they shrink geometrically. It's the resolution of Zeno's ancient paradox — Achilles really does catch the tortoise, because the infinitely many ever-shorter catch-up runs form a convergent geometric series with a perfectly finite sum.

Worked example — the bouncing ball's total journey. Back to our opening ball: dropped from 2 m, keeping 60\% of its height each bounce. It first falls 2 m. Then for every bounce it travels each height twice — up and back down — and the heights 1.2,\ 0.72,\ 0.432,\ \dots are geometric with a = 1.2, r = 0.6. Total distance:

Infinitely many bounces; exactly 8 metres. The promise from the top of the page, kept.

Worked example — why 0.999\dots = 1. The recurring decimal 0.999\dots is really a geometric series in disguise:

with a = \tfrac{9}{10} and r = \tfrac{1}{10}. So

Not "approximately 1", not "just under 1" — exactly 1. Two different decimal names for the same number, and the geometric series is the cleanest proof there is. (The same trick converts any recurring decimal into a fraction.)

See it explained