A geometric sequence multiplies by the same number at every step. The n-th term is a₁rⁿ⁻¹, and the sum of any finite number of terms comes from one line of cancellation. The exceptional ratio r = 1 has a simpler sum, and zero terms or a zero ratio are handled by the same multiplicative definition.
Core ideas
The formulas are the arithmetic ones with multiplication in place of addition. The cancellation has one excluded denominator, and that case is evaluated directly rather than discarded.
r = a₂/a₁, checked against a₃/a₂ and a₄/a₃, whenever those denominators are nonzero. Reaching for a subtraction here is the reflex left over from “Arithmetic Sequences and Series”: 3, 6, 12, 24 has differences 3, 6, 12 and ratios 2, 2, 2, and only the second list being constant makes it geometric. If zeros appear, check the multiplication aₙ₊₁ = raₙ directly instead.
a₁ has been multiplied by r no times, a₂ once, aₙ exactly n - 1 times. The exponent counts steps, and a₁rⁿ is the term after the one you wanted.
The cancellation formula works for a growing or alternating sequence as long as r ≠ 1. When r = 1, every term is a₁ and direct counting gives Sₙ = na₁. When r = 0, the sequence is a₁, 0, 0, … and the usual formula works without a special case. Convergence is “Infinite Geometric Series”'s question.
The material
The n-th term, the sum of n terms by cancellation, and the edge cases a quotient-only definition would miss.
A sequence with a first term a₁ and a constant multiplier r, called the common ratio, such that aₙ = r·aₙ₋₁ for n ≥ 2. Equivalently, aₙ = a₁rⁿ⁻¹. When consecutive terms in the denominator are nonzero, dividing aₙ₊₁/aₙ is the quickest way to find and check r; the multiplicative definition still makes sense when a term or r is zero.
3, 6, 12, 24, … has r = 2. 81, 27, 9, 3, … has r = 1/3 and shrinks. 5, -10, 20, -40, … has r = -2, and a negative ratio is how a sequence alternates in sign: odd-numbered terms keep the sign of a₁ and even-numbered ones flip it. 7, 0, 0, … has r = 0; after the first step the quotient 0/0 is unavailable, but every term still equals zero times the one before it.
aₙ = a₁rⁿ⁻¹ for every n ≥ 1. Getting from a₁ to aₙ is n - 1 multiplications by r, exactly as getting there in an arithmetic sequence was n - 1 additions of d.
For r > 0, the plotted terms lie on an exponential curve: a₁rⁿ⁻¹ is Exponential and Logarithmic Functions's a·bˣ sampled at the integers. For r < 0, no real exponential curve has that negative base; the magnitudes follow |a₁||r|ⁿ⁻¹ while the signs alternate. |r| > 1 grows in magnitude, 0 < |r| < 1 decays in magnitude, r = 1 is constant, and r = 0 becomes zero after the first term.
The sum of the terms of a geometric sequence. Sₙ denotes the sum of the first n terms, a₁ + a₁r + … + a₁rⁿ⁻¹.
For r ≠ 1, Sₙ = a₁(1 - rⁿ)/(1 - r). For r = 1 every term equals a₁ and Sₙ = na₁.
Write Sₙ = a₁ + a₁r + a₁r² + … + a₁rⁿ⁻¹, and multiply the whole line by r: rSₙ = a₁r + a₁r² + … + a₁rⁿ⁻¹ + a₁rⁿ.
Subtract the second line from the first. Every term of rSₙ also appears in Sₙ except the last, and every term of Sₙ appears in rSₙ except the first, so all but two cancel: Sₙ - rSₙ = a₁ - a₁rⁿ.
Factor both sides: Sₙ(1 - r) = a₁(1 - rⁿ). Dividing by 1 - r requires r ≠ 1, which is precisely the case the theorem handles separately, and there the sum is na₁ by inspection.
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Two edge cases are checks on the derivation rather than exceptions to the sequence. If r = 0, the formula gives Sₙ = a₁(1 - 0ⁿ)/(1 - 0) = a₁, matching a₁ + 0 + … + 0. If r = 1, the line Sₙ(1 - r) = a₁(1 - rⁿ) reduces to 0 = 0 and contains no sum information, so direct counting supplies Sₙ = na₁. Whether a sequence of finite sums approaches a number as n grows is a different question, taken up after partial sums have been made into a sequence of their own.
Interactive
The nth term and each finite partial sum computed from the definition and by formula
Method
Four steps. None of them asks whether an infinite sum exists; this lesson's stopping place is finite.
Divide at least twice: a₂/a₁ and a₃/a₂. Write r as an exact fraction, not a decimal — r = 1/3 raised to the tenth power is exact, and 0.333 raised to the tenth power is not.
Compute the power first, then multiply by a₁. With a negative r, settle the sign by the parity of the exponent before touching the arithmetic.
Both 1 - rⁿ and 1 - r may be negative; a negative divided by a negative is what makes a growing series come out positive.
When r = 1, use Sₙ = na₁ instead of dividing by 1 - r. In every case add the first few terms and compare their size and sign with the closed form; a finite sum is valid whether the terms shrink or grow.
One sequence, two questions: a distant term and a finite sum, each checked against the terms generated by the definition.
The check by hand is worth doing once. Ten terms added directly agree with the formula exactly, which verifies this one instance — a₁ = 3, r = 2, n = 10. What makes the formula true for every a₁, every r ≠ 1 and every n is the Sₙ - rSₙ cancellation, not the check.
For a faded geometric pass, use 81, -27, 9, -3, …. The supplied classification evidence is (-27)/81 = 9/(-27) = -1/3. Find a₈ and S₈ from the closed forms, then generate and add the eight terms directly as an independent check. Settle the sign of each power before multiplying its magnitude, then open the quiz.
You should be able to find or verify r, produce any term, sum any finite number of terms, handle r = 0 and r = 1 correctly, and reproduce the Sₙ - rSₙ cancellation in three lines. No convergence claim belongs in the answer yet.
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