How this instrument works
A geometric series adds the terms of a sequence that grows by a constant ratio rather than a constant difference — each term is the one before it multiplied by r, not increased by a fixed amount. That single change separates it sharply from an arithmetic series (the sum-of-linear-number-sequence sheet elsewhere on this site), where terms sit evenly spaced and the total is just n times the average of the first and last term. Here the terms compound, so the running total accelerates instead of climbing along a straight line.
The closed-form sum falls out of a classic trick: write S as the string of n terms, multiply that whole line by r, and subtract the two. Every middle term cancels, leaving S − rS = a₁ − a₁rⁿ, which rearranges directly into S = a₁(1 − rⁿ) ⁄ (1 − r). The trick works for any ratio except r = 1, where the subtraction divides by zero — but that case needs no formula at all, since every term is identical and the sum is simply a₁ multiplied by n.
Push n toward infinity and something unexpected shows up when the size of r stays under 1: the term rⁿ shrinks toward zero, so the sum settles on a finite limit, a₁ ⁄ (1 − r), even though infinitely many nonzero terms are being added — the resolution behind Zeno's runner, who covers half the remaining distance forever yet still finishes the race. Push r above 1 instead, as in a doubling sequence, and the running total grows almost as fast as the last term alone.
- Enter the starting number of your sequence into First term (a₁) — positive, negative, or a fraction all work.
- Enter the multiplier into Common ratio (r) — how much each term is scaled by to produce the next one.
- Enter the term count into Number of terms (n) — a positive whole number, counting the first term as term one.
- Read Sum of the series (total) — the exact running total of all n terms, built from the closed-form formula.
- If r comes out to exactly 1, expect a plain a₁ × n, since the formula's denominator would otherwise vanish.
Worked example — first term 3, ratio 2, five terms
Start a push-up habit at 3 repetitions on day one and double the count every day after that: First term (a₁) = 3, Common ratio (r) = 2, Number of terms (n) = 5. The five daily totals are 3, 6, 12, 24, and 48. Plugging straight into the formula gives S = 3(1 − 2⁵) ⁄ (1 − 2) = 3(1 − 32) ⁄ (−1) = 3 × (−31) ⁄ (−1) = 93, matching Sum of the series exactly.
Adding the five numbers by hand confirms it: 3 + 6 + 12 + 24 + 48 = 93. Notice the pattern specific to a ratio of 2: twice the final term minus the first term also gives 2 × 48 − 3 = 93, because a₁(2ⁿ − 1) always factors that way when r = 2. By day five the single day's count, 48, is already more than half the five-day total.
Questions
What is the difference between a geometric series and an arithmetic series?
A geometric series multiplies each term by a constant ratio r; an arithmetic series adds a constant difference instead. Because geometric terms compound, their sum needs a formula built from powers of r, S = a₁(1 − rⁿ) ⁄ (1 − r), while an arithmetic sum only needs the average of the first and last term times the number of terms.
How is the sum of a geometric series formula actually derived?
Write S as a₁ + a₁r + a₁r² + … + a₁rⁿ⁻¹, multiply every term of that line by r, and subtract the new line from the original. All the middle terms cancel, leaving S − rS = a₁ − a₁rⁿ, which rearranges directly into S = a₁(1 − rⁿ) ⁄ (1 − r).
What happens if the common ratio equals 1?
The formula's denominator becomes zero, so it cannot be used directly — but the fix is simple: when r = 1, every term equals a₁, so the sum is just a₁ multiplied by n. A first term of 5 repeated 4 times, for instance, sums to 20 with no rⁿ involved at all.
Does a geometric series always keep growing as more terms are added?
Only when the size of the ratio is 1 or greater. If r is smaller than 1 in size, each new term shrinks past the last, and the running total creeps toward a fixed ceiling, a₁ ⁄ (1 − r), even as n grows without bound — the same idea behind Zeno's runner covering ever-smaller remaining distances yet still reaching the finish.
What is a common mistake when counting the number of terms, n?
An off-by-one slip — treating the exponent on the last term as n instead of n − 1, or folding in an extra starting value that was never part of the sequence. Recount by listing the actual values (a₁, a₁r, a₁r², …) rather than trusting a remembered formula, then check that count against the last term you expect to see.
Is there a shortcut when the common ratio is exactly 2?
Yes — for r = 2 the sum always equals twice the final term minus the first term, S = 2 × (last term) − a₁, because a₁(2ⁿ − 1) factors that way. It is the same doubling arithmetic behind the old chessboard-and-wheat legend, where grain counts square by square outpace an entire kingdom's granary.