In short

The sum of the first n terms of a geometric progression with first term a and common ratio r \neq 1 is S_n = a\,\dfrac{r^n - 1}{r - 1}. When |r| < 1 and you keep adding forever, the terms shrink to nothing and the infinite sum converges to S_\infty = \dfrac{a}{1 - r}. Both formulas come from the same trick: multiply the sum by r, shift everything by one position, and subtract to cancel almost every term.

A ball is dropped from a height of 18 metres. Each time it hits the ground, it bounces back to \dfrac{2}{3} of the height it fell from. So it rises to 12 metres after the first bounce, then 8 metres, then \dfrac{16}{3} metres, and so on. You want to know: what is the total distance the ball travels before it comes to rest?

The ball falls 18 metres, rises 12, falls 12, rises 8, falls 8, and so on. The upward distances alone form a GP: 12, 8, \dfrac{16}{3}, \dfrac{32}{9}, \dots with first term 12 and common ratio \dfrac{2}{3}. Each bounce is smaller than the last, and the total distance is finite — the GP converges. To find that total, you need a formula for summing the terms of a GP. That is what this article derives.

The finite sum formula

The trick: multiply and subtract

Take the GP a, ar, ar^2, \dots, ar^{n-1} and call its sum S_n:

S_n = a + ar + ar^2 + \cdots + ar^{n-1}

Now multiply both sides by r:

rS_n = ar + ar^2 + ar^3 + \cdots + ar^n

Look at the two lines. The right-hand side of the second line is almost identical to the first — every term of S_n except the first appears in rS_n, and rS_n has one extra term (ar^n) at the end. Subtract the first equation from the second:

rS_n - S_n = ar^n - a
(r - 1)\,S_n = a(r^n - 1)
\boxed{S_n = a\,\frac{r^n - 1}{r - 1}} \qquad (r \neq 1)

That is it. One multiplication, one subtraction, and all the intermediate terms cancel in pairs. The formula works for any r \neq 1. When r = 1, every term equals a and the sum is just na — no formula needed.

Visual derivation of the GP sum formula by shifting and subtractingTwo rows of boxes representing the terms of S sub n and r times S sub n. The first row has boxes labelled a, ar, ar squared, dot dot dot, ar to the n minus 1. The second row, shifted one position to the right, has boxes labelled ar, ar squared, ar cubed, dot dot dot, ar to the n. Arrows show that all middle terms cancel when subtracting, leaving only a on the left of the first row and ar to the n on the right of the second row. S n = a ar ar² ... ar ⁿ⁻¹ rS n = ar ar² ar³ ... ar ⁿ cancel cancel subtract: −a + 0 + 0 + ··· + 0 + +ar ⁿ (r − 1) · Sₙ = a(rⁿ − 1)
The multiply-and-subtract trick. The top row is $S_n$; the second row is $rS_n$, shifted one position to the right. The dashed red lines show terms that appear in both rows and cancel when you subtract. Only the first term of $S_n$ ($a$, with a sign flip) and the last term of $rS_n$ ($ar^n$) survive, giving $(r - 1)S_n = a(r^n - 1)$.

The two forms of the formula

Depending on whether r > 1 or r < 1, one form of the formula is more convenient than the other:

S_n = a\,\frac{r^n - 1}{r - 1} \qquad \text{(natural when } r > 1 \text{, so both numerator and denominator are positive)}
S_n = a\,\frac{1 - r^n}{1 - r} \qquad \text{(natural when } r < 1 \text{, same reason)}

These are algebraically identical — multiply the top and bottom of either by -1 to get the other. Use whichever keeps the signs clean.

The infinite GP sum

What happens when |r| < 1

Take the finite sum formula and ask what happens as n \to \infty:

S_n = a\,\frac{1 - r^n}{1 - r}

When |r| < 1, the term r^n shrinks as n grows. For example, \left(\tfrac{1}{2}\right)^{10} = \tfrac{1}{1024} \approx 0.001, and \left(\tfrac{1}{2}\right)^{20} \approx 0.000001. As n increases without bound, r^n \to 0, and the sum approaches

\boxed{S_\infty = \frac{a}{1 - r}} \qquad (|r| < 1)

This is the sum to infinity of a GP. The infinite sum is finite because the terms shrink fast enough that their total cannot grow without bound.

Partial sums of the GP with a equals 8 and r equals one half approaching 16A coordinate plane with partial sum index n on the horizontal axis from 1 to 10 and partial sum value on the vertical axis from 0 to 18. A horizontal dashed red line marks the infinite sum at 16. Ten dots show the partial sums: 8, 12, 14, 15, 15.5, and so on, approaching 16 from below. Each successive dot is closer to the red line, but none touches it. 0 8 16 S∞ = 16 1 2 3 4 5 6 7 8 9 8 12 14 15 n
The partial sums of the GP $8 + 4 + 2 + 1 + \cdots$ ($a = 8$, $r = \tfrac{1}{2}$). Each dot is the sum of the first $n$ terms. The dashed red line at $S_\infty = \dfrac{8}{1 - 0.5} = 16$ is the limit the partial sums approach but never reach. By $n = 9$, the sum is already within $0.03$ of $16$.

When |r| \geq 1, the term r^n does not shrink. If r = 1, every term is a and the sum grows without bound. If r = -1, the sum oscillates (a, 0, a, 0, \dots) without settling. If |r| > 1, the terms themselves grow, and the sum diverges. The infinite GP sum exists only when |r| < 1.

Visualising the infinite sum with stacking

There is a beautiful geometric way to see why 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots = 2. Start with a square of side length 1 (area = 1). Place a rectangle of area \dfrac{1}{2} next to it. Then a rectangle of area \dfrac{1}{4}, then \dfrac{1}{8}, and so on. Each piece is half the size of the previous one, and together they tile a rectangle of dimensions 2 \times 1 — total area 2.

The infinite GP sum 1 plus one half plus one quarter plus one eighth equals 2 shown as rectangles tiling a 2 by 1 stripA horizontal rectangle of width 2 units and height 1 unit. The left half is a square of area 1, labelled 1. The right half is subdivided: a tall rectangle of area one half on the left, then a square of area one quarter, then stacked rectangles of area one eighth and one sixteenth filling the remaining space. Each piece is half the area of the previous one. Together they exactly fill the two-by-one rectangle, confirming the sum equals 2. 1 1/2 1/4 1/4 1/8 ... total width = 2
The GP $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$ tiles a $2 \times 1$ rectangle exactly. Each piece is half the area of the previous one. No matter how many pieces you add, you never exceed the rectangle — and the pieces fill it completely. The sum is $\dfrac{1}{1 - 1/2} = 2$.

Applications

Application 1: Recurring decimals

The recurring decimal 0.\overline{36} = 0.363636\dots is an infinite GP in disguise:

0.36 + 0.0036 + 0.000036 + \cdots

The first term is a = 0.36 and the common ratio is r = 0.01 (each block is shifted two decimal places to the right). Since |r| = 0.01 < 1, the infinite sum formula applies:

S_\infty = \frac{0.36}{1 - 0.01} = \frac{0.36}{0.99} = \frac{36}{99} = \frac{4}{11}

This recovers the fraction \dfrac{4}{11} from its decimal expansion — the same result the multiply-and-subtract trick from Number Systems gives, but through a different lens. Every recurring decimal is an infinite GP, and the sum-to-infinity formula is the machinery that converts it back to a fraction.

Application 2: Compound interest

You invest ₹P at an annual interest rate of r_{\%} compounded yearly. The balance at the end of year n is P(1 + r_{\%}/100)^n — a GP term. But suppose you also deposit ₹D at the end of every year. What is the total balance after n years?

Each deposit earns interest for a different number of years. The deposit made at the end of year 1 grows for n - 1 years to become D(1 + r_{\%}/100)^{n-1}. The deposit at the end of year 2 grows for n - 2 years. And the last deposit (year n) earns no interest.

The total from the deposits is

D + D(1 + r_{\%}/100) + D(1 + r_{\%}/100)^2 + \cdots + D(1 + r_{\%}/100)^{n-1}

This is a finite GP with first term D, common ratio (1 + r_{\%}/100), and n terms. By the sum formula:

\text{total deposits} = D \cdot \frac{(1 + r_{\%}/100)^n - 1}{r_{\%}/100}

This is the formula banks use for Systematic Investment Plans (SIPs) and recurring deposits — a direct application of the GP sum.

Application 3: Bouncing ball

Return to the opening example. The ball is dropped from 18 metres and bounces to \dfrac{2}{3} of its previous height each time. The total distance is:

\text{distance} = 18 + 2(12 + 8 + \tfrac{16}{3} + \cdots)

The factor of 2 accounts for both the rise and the fall of each bounce. The infinite GP in the brackets has a = 12 and r = \dfrac{2}{3}:

12 + 8 + \tfrac{16}{3} + \cdots = \frac{12}{1 - 2/3} = \frac{12}{1/3} = 36

So the total distance is 18 + 2 \times 36 = 18 + 72 = 90 metres.

Bouncing ball heights forming a GP with ratio two thirdsA sequence of vertical bars representing the bounce heights of a ball dropped from 18 metres. The first bar is 18 units tall, representing the initial drop. The next bar is 12 units tall (the first bounce), then 8, then 16 thirds which is about 5.3, then about 3.6, each bar two thirds the height of the previous one. A dashed red line at the top of each bar connects the peaks, showing the exponential decay. 0 18 12 8 drop b1 b2 b3 b4 ... 18 12 8
Each bounce reaches $\dfrac{2}{3}$ of the previous height. The dashed red curve through the peaks traces out a GP. The total vertical distance (all rises and falls combined) is $18 + 2 \times 36 = 90$ metres — a finite answer from infinitely many bounces.

Two worked examples

Example 1: Find the sum of the first 8 terms of the GP $3, 6, 12, 24, \dots$

Step 1. Identify a and r.

a = 3, \qquad r = \frac{6}{3} = 2

Why: the first term is 3 and the common ratio is found by dividing consecutive terms.

Step 2. Choose the appropriate form of the formula. Since r = 2 > 1, use

S_n = a\,\frac{r^n - 1}{r - 1}

Why: with r > 1, both the numerator and denominator come out positive, so the signs stay clean.

Step 3. Compute r^n.

2^8 = 256

Why: the exponent is n = 8, the number of terms. You can build this up: 2^4 = 16, so 2^8 = 16^2 = 256.

Step 4. Substitute into the formula.

S_8 = 3 \cdot \frac{256 - 1}{2 - 1} = 3 \cdot \frac{255}{1} = 765

Result: S_8 = 765.

Stacked bars showing the 8 terms of the GP summing to 765Eight horizontal bars stacked vertically, each representing one term of the GP. The widths are proportional to the term values: 3, 6, 12, 24, 48, 96, 192, 384. Each bar is labelled with its value. A total of 765 is shown at the right. 3 6 12 24 48 96 192 384 total = 3 + 6 + 12 + ··· + 384 = 765
Each term of the GP is twice the previous one, so each bar is twice as wide. The last term ($384$) alone is more than half the total sum — this is typical of GPs with $r > 1$, where the bulk of the sum is concentrated in the last few terms.

The last term (384) is more than half the total (765). This is a characteristic feature of GPs with r > 1: the latest terms dominate the sum, because each term is larger than all the previous terms combined (in fact, for r = 2, each term equals the sum of all previous terms plus a).

Example 2: Find the sum to infinity of the GP $27, 9, 3, 1, \dfrac{1}{3}, \dots$

Step 1. Identify a and r.

a = 27, \qquad r = \frac{9}{27} = \frac{1}{3}

Why: each term is one-third of the previous one.

Step 2. Check that |r| < 1.

\left|\frac{1}{3}\right| = \frac{1}{3} < 1 \quad ✓

Why: the infinite sum formula only applies when the terms are shrinking toward zero, which requires |r| < 1.

Step 3. Apply the infinite sum formula.

S_\infty = \frac{27}{1 - 1/3} = \frac{27}{2/3} = 27 \times \frac{3}{2} = \frac{81}{2} = 40.5

Why: dividing by a fraction is the same as multiplying by its reciprocal.

Step 4. Verify with partial sums. S_1 = 27. S_2 = 36. S_3 = 39. S_4 = 40. S_5 = 40.\overline{3}. The sums are indeed climbing toward 40.5 from below.

Result: S_\infty = \dfrac{81}{2} = 40.5.

Partial sums of 27 plus 9 plus 3 plus 1 plus one third plus dot dot dot approaching 40 point 5A coordinate plane with term count n on the horizontal axis from 1 to 8 and partial sum on the vertical axis from 0 to 45. A dashed red horizontal line at 40.5 marks the infinite sum. Eight dots show the partial sums climbing from 27 toward the red line, approaching it rapidly. 0 40 20 S∞ = 40.5 1 2 3 4 5 6 7 27 36 39 40
The partial sums climb rapidly toward $40.5$. By the 4th partial sum, you are already at $40$ — within $1.2\%$ of the infinite total. This fast convergence is typical when $|r|$ is small: most of the sum is captured by the first few terms.

The speed of convergence depends on |r|. When |r| is close to 0 (as here, r = \tfrac{1}{3}), the partial sums race toward the limit. When |r| is close to 1, convergence is painfully slow.

Common confusions

Going deeper

If you came here to learn the finite and infinite GP sum formulas and see how they apply, you have them. The rest of this section is for readers who want to push into two directions: a deeper look at why the infinite sum converges, and a connection to a beautiful piece of mathematics.

Why does the infinite sum converge?

The formula tells you that S_\infty = \dfrac{a}{1 - r} when |r| < 1, but why is a sum of infinitely many positive numbers finite? The answer comes from looking at what is left over.

After n terms, the partial sum is S_n = a \cdot \dfrac{1 - r^n}{1 - r}. The "remainder" — everything you have not yet added — is

R_n = S_\infty - S_n = \frac{a}{1 - r} - a \cdot \frac{1 - r^n}{1 - r} = \frac{a \cdot r^n}{1 - r}

This remainder is a GP term scaled by \dfrac{1}{1 - r}. Since |r| < 1, the term r^n shrinks exponentially. After 10 terms of the GP with r = \tfrac{1}{2}, the remainder is \dfrac{a}{1024(1 - r)} — a thousandth of the full sum. After 20 terms, it is a millionth. The remainder vanishes not because you run out of terms, but because each remaining term is so small that even infinitely many of them cannot add up to anything significant.

This is the heart of convergence: the tail of the series is negligible, not because the terms are zero, but because they shrink so fast that their total can be made as small as you like.

Connection to \dfrac{1}{1 - x}

The finite GP sum can be written as

1 + x + x^2 + \cdots + x^{n-1} = \frac{1 - x^n}{1 - x}

(take a = 1, r = x). When |x| < 1 and n \to \infty, the left side becomes the infinite series

1 + x + x^2 + x^3 + \cdots = \frac{1}{1 - x}

This identity — the geometric series — is one of the most important formulas in mathematics. It expresses a simple rational function \dfrac{1}{1 - x} as an infinite sum of powers. The idea that a function can be written as an infinite polynomial (a power series) is the starting point of Taylor series, Fourier series, and much of analysis. Every one of those topics begins by generalising the geometric series to functions other than \dfrac{1}{1 - x}.

The GP sum formula is where this entire story starts.

The Koch snowflake's perimeter

Here is a famous application. Start with an equilateral triangle of side length 1. On each side, build a smaller equilateral triangle on the middle third. Repeat forever. The result is the Koch snowflake — a curve that encloses a finite area but has infinite perimeter.

At each step, the number of sides is multiplied by 4 and each side is divided by 3. The perimeter at step n is 3 \cdot \left(\dfrac{4}{3}\right)^n. Since \dfrac{4}{3} > 1, this GP diverges — the perimeter grows without bound. But the area added at each step forms a GP with ratio \dfrac{4}{9} < 1, so the total area converges. A shape with finite area and infinite boundary — geometry built from the convergence and divergence of GPs.

Where this leads next

The GP sum formula opens the door to infinite series, one of the central ideas of analysis and calculus.