Geometric Sequence
A geometric sequence multiplies by a fixed common ratio r. This calculator finds the nth term aₙ = a₁·r^(n−1) and the partial sum Sₙ = a₁(1 − rⁿ)/(1 − r). When |r| < 1 it also reports the infinite sum S∞ = a₁/(1 − r).
One term, a partial sum, and sometimes a limit
A geometric sequence multiplies rather than adds. Each term is the previous one times a fixed number, the common ratio, so terms grow or shrink by a constant proportion rather than a constant amount. That change from addition to multiplication is what makes the behaviour dramatic: doubling repeatedly outruns any amount of adding.
This calculator takes the first term, the ratio and a term number, and returns that term, the sum of the terms up to it, and — when the ratio is small enough for the series to settle — the total of infinitely many of them.
The three results answer different questions. The nth term asks where the sequence has got to; the partial sum asks how much has accumulated along the way; the infinite sum asks whether that accumulation ever stops growing.
Only the third is conditional, and the condition is sharp. A ratio smaller than one in magnitude makes the terms shrink fast enough that the total converges; anything else and the sum grows without bound. The tool reports the infinite sum only when that holds, rather than printing a meaningless number.
How to use this calculator
- Enter the first term The value the sequence starts from. Any real number, including negatives and fractions. It scales everything else proportionally.
- Enter the common ratio The multiplier between consecutive terms. A ratio above one grows, between zero and one shrinks, and a negative ratio alternates in sign as it does either.
- Enter the term number A whole number of at least one, counting from the first term. A fractional or zero value is rejected with an explicit message.
- Read the infinite sum only when it appears That line is present exactly when the ratio is less than one in magnitude. Its absence is a result about the series, not a missing calculation.
The formula, and where it comes from
aₙ = a₁·r^(n−1) Sₙ = a₁(1 − rⁿ)/(1 − r) S∞ = a₁/(1 − r) for |r| < 1
The nth-term formula uses n − 1 rather than n because the first term has been multiplied by the ratio no times at all. Off-by-one errors here are the single most common mistake with geometric sequences, and the exponent is where they live.
The partial-sum formula comes from a short trick: write the sum, multiply it by r, subtract. Almost every term cancels, leaving the first and a copy of the last, and rearranging gives the quotient above. That also explains the division by 1 − r, and why the formula fails when the ratio is exactly one.
When the ratio is one, every term equals the first and the sum is just the first term times n. The implementation checks for that case separately rather than dividing by zero, which is why a ratio of one returns a clean answer instead of an error.
The infinite sum is the partial sum with rⁿ driven to zero, which happens precisely when the ratio is under one in magnitude. What remains is the first term divided by one minus the ratio — a finite total for infinitely many terms.
What each input means
- a₁ First term — form field “First term a₁”
- Where the sequence starts. Every reported quantity is proportional to it, so doubling it doubles the term, the partial sum and the infinite sum alike.
- r Common ratio — form field “Common ratio r”
- The constant multiplier between terms. Its magnitude decides growth or decay; its sign decides whether consecutive terms alternate.
- n Term number — form field “Term number n”
- Which term to report, counting the first as one. Must be a positive whole number.
- Sₙ, S∞ Sums
- The total of the first n terms, and the total of all of them. The second exists only when the ratio is under one in magnitude.
Worked examples
Every number below is produced by the same calculation engine the tool above runs. Nothing here is typed by hand, so the walkthrough cannot drift from what you get when you enter the same values yourself.
A sequence that grows
Start at 2 with a ratio of 3, and ask for the eighth term. Multiplying by three each time compounds quickly, so this shows the scale geometric growth reaches.
Inputs First term a₁ = 2, Common ratio r = 3, Term number n = 8
- Given a₁ = 2, common ratio r = 3, n = 8
- nth term aₙ = a₁·r^(n−1) = 2·3^7 = 4374
- Sum of n terms Sₙ = a₁·(1 − rⁿ)/(1 − r) = 6560
Result a₈ = 4374, Sₙ = 6560
The eighth term uses an exponent of seven, not eight. Substituting the term number directly gives a value three times too large, which is exactly the off-by-one the formula's n − 1 is guarding against.
No infinite sum is reported, and there could not be one. With a ratio above one the terms keep growing, so the running total has no limit to approach.
A series that converges
Start at 1 with a ratio of one half, over ten terms. Each term is half the last, so the sequence shrinks towards zero and the total approaches a limit.
Inputs First term a₁ = 1, Common ratio r = 0.5, Term number n = 10
- Given a₁ = 1, common ratio r = 0.5, n = 10
- nth term aₙ = a₁·r^(n−1) = 1·0.5^9 = 0.00195313
- Sum of n terms Sₙ = a₁·(1 − rⁿ)/(1 − r) = 1.99805
- Infinite sum |r| < 1 → S∞ = a₁/(1 − r) = 2
Result a₁₀ = 0.00195313, Sₙ = 1.99805, S∞ = 2
The partial sum after ten terms is already within a thousandth of the infinite sum of 2. Halving means each new term contributes half of what remains to be contributed, so the total closes in on its limit rapidly.
The infinite-sum line appears here because the ratio is under one in magnitude. It says that adding infinitely many positive numbers can still produce a finite total, provided they shrink fast enough.
Reading the result
What the ratio tells you at a glance
Above one in magnitude, the terms grow without bound and so does the sum. Below one, both the terms and the sum settle. Exactly one gives a constant sequence, and exactly minus one gives one that flips between two values forever without settling.
A negative ratio alternates
Odd-numbered terms keep the sign of the first and even-numbered ones flip it. The partial sums then oscillate above and below their limit rather than climbing towards it — worth expecting before reading a sum that looks too small.
When you would use this
Compound growth term by term
A balance earning a fixed rate, a population growing by a fixed percentage, a quantity decaying by a fixed fraction: all geometric, and the nth term says where the quantity stands after n periods.
Converting a repeating decimal to a fraction
A repeating decimal is a geometric series with a ratio of a tenth, a hundredth or similar. The infinite-sum formula turns it into an exact fraction, which is the standard proof that every repeating decimal is rational.
Assumptions and limitations
What this calculator assumes
- The sequence is indexed from one, so the first term corresponds to n = 1.
- The term number is a positive whole number; anything else is rejected rather than rounded.
- The ratio of exactly one is handled by a separate formula, since the general partial-sum expression would divide by zero.
- The infinite sum is reported only when the ratio is strictly less than one in magnitude.
Where it stops being the right tool
- The first term and ratio must be known: they are not recovered from two given terms of an unknown sequence.
- No arithmetic or mixed sequences — a sequence with a constant difference is a different tool.
- Only forward indexing from the first term, so a sum starting from some later term must be found by subtracting two partial sums.
Common mistakes
Using n instead of n − 1 in the exponent
Why it happens. The term is numbered n, so raising the ratio to n looks consistent. The first term is the ratio to the power zero, which shifts every exponent down by one.
How to avoid it. Test with n = 1. The formula must return the first term unchanged; if yours returns the first term times the ratio, the exponent is one too high.
Expecting an infinite sum for a growing series
Why it happens. The partial sum is a finite number for every n, so it is tempting to assume that continuing forever gives some large but finite total.
How to avoid it. Check the magnitude of the ratio. At one or above the terms do not shrink, the partial sums grow without limit, and no infinite sum exists.
Confusing the nth term with the sum to n terms
Why it happens. Both are reported together and both grow with n, so the wrong line can be quoted without anything looking obviously amiss.
How to avoid it. Compare the two: the sum is always the larger for a positive ratio, since it includes that term and every one before it.
Frequently asked questions
What is the nth term formula?
aₙ = a₁·r^(n−1). The exponent is one less than the term number because the first term has not yet been multiplied by the ratio at all.
When does the infinite sum exist?
Only when the ratio is strictly less than one in magnitude, giving S∞ = a₁/(1 − r). At that point the terms shrink fast enough for the total to settle on a finite value; at a ratio of one or more they do not.
What happens if the ratio is exactly 1?
Every term equals the first, so the sum of n terms is simply the first term times n. The calculator handles that case separately, since the general formula would divide by zero.
Can the ratio be negative?
Yes. The terms then alternate in sign, and the partial sums oscillate around their eventual value rather than approaching it from one side. Convergence still depends only on the magnitude of the ratio.