What makes a sequence geometric?

Each term is the previous one multiplied by a fixed number. Divide each term by the one before it: if the answer is always the same, that number is the common ratio. Testing three or four consecutive ratios is usually enough. The contrast with an arithmetic sequence is in the operation rather than the pattern — one multiplies by a constant, the other adds one, and checking both takes only a moment.

Why is the exponent n − 1 rather than n?

Because the first term has not been multiplied yet. Reaching the second term takes one multiplication, the third takes two, and the nth takes n − 1. Using rⁿ gives a₁r^n, which is the term after the one you wanted. It is the same fencepost problem as in arithmetic sequences: there are always one fewer operations than terms, because the first term is where you start rather than somewhere you arrive.

When does an infinite geometric series have a sum?

Only when the common ratio lies strictly between −1 and 1. Inside that band each term is smaller in size than the last, the amounts being added shrink towards nothing, and the running total settles on a₁ ÷ (1 − r). At |r| = 1 the terms never shrink — they either stay constant or flip sign forever — and beyond it they grow, so in both cases the total has nowhere to settle.

Why does 0.999… equal exactly 1?

Because it is a geometric series with a finite total. Written out it is 0.9 + 0.09 + 0.009 + …, which has a₁ = 0.9 and r = 0.1, and the infinite sum formula gives 0.9 ÷ (1 − 0.1) = 0.9 ÷ 0.9 = 1. There is no approximation and no limit being waved at — the notation 0.999… means the value of that series, and that value is 1. The same argument turns any recurring decimal into a fraction.

Can the common ratio be negative?

Yes, and the terms then alternate between positive and negative at every step. A ratio of −2 gives 1, −2, 4, −8, 16 — the sizes still double, but the sign flips each time. Convergence depends only on the size of the ratio, not its sign, so a ratio of −0.5 gives a perfectly well-behaved infinite total. The partial sums oscillate above and below the limit as they close on it rather than approaching from one side.

Why does finding the ratio sometimes give two answers?

Because an even root has two real values. If the two known terms are an even number of positions apart, the ratio between them is an even power of r, and both a positive and a negative r produce it. From a₁ = 1 and a₃ = 4 the ratio squared is 4, so r is 2 or −2, and both give sequences that match at every position you specified. An odd gap gives exactly one answer, and an even gap with a negative ratio between the terms gives none at all, since no real number raised to an even power is negative.

What happens when the common ratio is 1?

Every term equals the first and the sequence is constant. The sum formula cannot be used because it divides by 1 − r, which is zero, so a calculator applying it blindly returns nothing usable. The correct total is simply n × a₁. This is also the only ratio at which a sequence is both geometric and arithmetic, since multiplying by one and adding zero have the same effect.

Can a geometric sequence contain zero?

No. If any term were zero then every later term would be zero too, since multiplying zero by anything leaves zero — and the ratio between two consecutive zeros would be 0 ÷ 0, which is not a number. So both the first term and the common ratio must be non-zero. The sequence 0, 0, 0, … is perfectly well defined as a list; it simply has no common ratio, so none of the geometric formulas apply to it.

How quickly does geometric growth actually get out of hand?

Faster than almost anyone expects. Doubling a single grain of wheat once per square of a chessboard reaches 2⁶⁴ − 1 grains in total, which is more than eighteen quintillion — vastly more wheat than has ever been grown. The practical consequence is that any geometric sequence with a ratio above 1 eventually overtakes any arithmetic sequence, however large the arithmetic one’s common difference, and once it does it never falls behind again.

Why do the terms stop being exact whole numbers?

Because a double-precision number can hold consecutive integers only up to 2⁵³, roughly nine quadrillion. Past that the gaps between representable values exceed one, so a term shown as a whole number is really the nearest value the format can store. With a ratio of 2 this happens at the 53rd term — and yet the values do not overflow to infinity until the 1024th. Nearly a thousand terms sit in between looking like exact integers while quietly not being any.

Geometric Sequence Calculator

A geometric sequence multiplies by the same amount each step, so aₙ = a₁ · r^(n−1) and the first n terms add to a₁(1 − rⁿ) ÷ (1 − r). Three things here have no counterpart in an arithmetic sequence. An infinite geometric series can have a finite total, but only when |r| < 1. That is why 0.999… is ...

An infinite geometric series can have a finite total — but only when |r| < 1. That is not a curiosity: it is why 0.999… equals exactly 1. Written out it is 0.9 + 0.09 + 0.009 + …, a geometric series with a₁ = 0.9 and r = 0.1, and 0.9 ÷ (1 − 0.1) is 1 with no approximation anywhere.

THE SEQUENCE

a₁r

nhow many terms

aₙ = 3 × 2^{(n − 1)}

a10

1536

S10

3069

a10 is 3 multiplied by 2 9 times, not 10. Because |r| is above 1, no infinite total exists — the terms never shrink towards nothing.

The terms. 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536

Quantity	Working	Result
Multiplications	n − 1 = 10 − 1	9
a10	3 × 2^9	1536
S10	3(1 − 2^10) ÷ (1 − 2)	3069
S∞	\|r\| ≥ 1, so none exists	—
Ratio check	a2 ÷ a1 = 6 ÷ 3	2

THE SHAPE OF IT

Multiplying repeatedly changes the scale, not just the size. An arithmetic sequence adds the same amount each step and traces a straight line; a geometric one multiplies, so the steps themselves grow or shrink. Given any ratio above 1, a geometric sequence eventually overtakes every arithmetic one, however large its common difference.

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HOW TO USE

1
Give the first term and the common ratio, or give any two terms. Where two terms are an even number of positions apart the page offers both ratios rather than silently picking the positive one.
2
Watch the ratio marker against −1 and 1. Inside that band an infinite total exists; on or outside it the terms never shrink towards nothing and the sum runs away.
3
Read any warning about precision seriously. A term shown as a whole number may not be one — beyond 2⁵³ a double cannot hold consecutive integers, and the figure displayed is the nearest available value.
4
Remember that neither the first term nor the ratio may be zero. One zero term makes every later term zero and the ratio between them undefined.

REFERENCE FORMULAS

Rule	Formula	What it is for
nth term	aₙ = a₁ · r^(n − 1)	The exponent is n − 1, not n. The first term has been multiplied zero times.
Common ratio	r = aₙ₊₁ ÷ aₙ	Constant by definition. If consecutive ratios differ, the sequence is not geometric.
From two terms	r = (aₘ ÷ aₖ)^(1/(m − k))	An even gap gives two real answers, ±. An odd gap gives exactly one.
Sum of n terms	Sₙ = a₁(1 − rⁿ) ÷ (1 − r)	Only for r ≠ 1. At r = 1 this is 0 ÷ 0 and the sum is simply n·a₁.
Infinite sum	S∞ = a₁ ÷ (1 − r), \|r\| < 1	Diverges otherwise. At \|r\| ≥ 1 the terms do not shrink fast enough to settle.
Recurring decimal	0.999… = 0.9 ÷ (1 − 0.1) = 1	Exactly one, not nearly. Any recurring decimal is a geometric series in disguise.
Negative ratio	r < 0 alternates sign	Terms swing either side of zero. The sum still converges if \|r\| < 1.
Geometric mean	√(aₖ · aₖ₊₂) = \|aₖ₊₁\|	The middle of three consecutive terms, up to sign. Not the arithmetic mean.
Doubling	r = 2 ⟹ aₙ = a₁ · 2^(n−1)	The chessboard problem: 64 squares reach 2⁶⁴ − 1 grains in total.
Zero forbidden	a₁ ≠ 0 and r ≠ 0	One zero term makes every later term zero, and the ratio 0 ÷ 0 is undefined.
Product of terms	a₁ⁿ · r^(n(n−1)/2)	The exponents 0, 1, 2 … n−1 add to a triangular number.
Growth against arithmetic	geometric overtakes any arithmetic	Given r > 1, eventually and permanently, however large the common difference.
Exact to	r = 2 stays exact to n = 53	Past 2⁵³ a double cannot hold consecutive integers. Overflow is much later, at n = 1024.
Not arithmetic	geometric multiplies; arithmetic adds	Test ratios for one, differences for the other. Only a constant sequence is both.

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Last updated: August 4, 2026 · aₙ = a₁·r^(n−1) · An infinite total exists only when |r| < 1, and two terms an even distance apart admit two ratios.