Multiply by the same ratio every time, and the running total creeps toward a limit.
Take a number and keep multiplying by the same factor. 2, 6, 18, 54, … triples at every step; 16, 8, 4, 2, … halves at every step. Where an arithmetic sequence adds a constant and walks up a straight line, a geometric sequence multiplies by a constant and rides an exponential curve. That one change — add becomes multiply — turns the term formula into a power of q, lets growth explode or decay toward zero, and produces something an arithmetic sequence never could: an endless sum that still settles on a finite answer. By the end you will know exactly when "add forever" stops at a number, and why Zeno's 12 + 14 + 18 + ⋯ equals exactly 1.
A sequence {aₙ} is geometric when every term is the one before it times the same number. That fixed multiplier is the common ratio q:
q = aₙ₊₁aₙ (the same value for every n).
To test a sequence, divide each term by the one before. In 2, 6, 18, 54: 6÷2 = 3, 18÷6 = 3, 54÷18 = 3 — every ratio is 3, so q = 3 and the sequence triples. In 16, 8, 4, 2: 8÷16 = ½, 4÷8 = ½, 2÷4 = ½ — every ratio is ½, so q = ½ and the sequence halves. Because we divide by aₙ, no term may be 0, so we require a₁ ≠ 0 and q ≠ 0.
A sequence is geometric exactly when the ratio of consecutive terms is constant: q = aₙ₊₁ / aₙ. Arithmetic asks "what did you add?" (a constant difference d); geometric asks "what did you multiply by?" (a constant ratio q).
2, 6, 18, 54 has differences 4, 12, 36 — not constant, so it is not arithmetic. But its ratios are all 3, so it is geometric. Check the right thing: arithmetic lives in subtraction, geometric in division.
To reach the n-th term from a₁, you multiply by q a total of n−1 times — once to get to a₂, again to get to a₃, and so on. So:
aₙ = a₁ · qn−1
For 2, 6, 18, 54 with a₁ = 2, q = 3: a₄ = 2 · 3³ = 2 · 27 = 54 ✓. Notice the exponent is n−1, not n — the first term has taken zero steps, and q⁰ = 1.
Look at that formula as a function of n. It is a constant times q raised to the power n — exactly the shape of an exponential function. So the dots of a geometric sequence sample an exponential curve: when |q| > 1 the terms blow up, and when |q| < 1 they decay toward 0 without ever reaching it. A constant difference gave us a line in 26.2; a constant ratio gives us an exponential here.
Dial the first term and the ratio. Watch the dots curve up (|q|>1), decay to 0 (|q|<1), or flip sign (q<0). The readout gives aₙ, S₅, and the infinite sum (or "diverges").
The first term of a geometric sequence is 3 and the ratio is 2. Find the 8th term. a₈ = 3 · 2⁷ = 3 · 128 = 384. Multiplying by 2 seven times turns 3 into 384 — exponential growth is fast.
In an arithmetic sequence, the middle of three in a row is the average of its neighbours (their sum halved). In a geometric sequence, the middle is the geometric mean — the multiplicative middle. If a, b, c are three consecutive terms, then the ratio is the same on both sides:
ba = cb ⟹ b² = ac ⟹ b = ±√(ac).
So b is a geometric mean of a and c exactly when b² = ac. For 2 and 8: b² = 2·8 = 16, so b = ±4 — and indeed 2, 4, 8 is geometric with q = 2, while 2, −4, 8 works too with q = −2. The geometric mean exists (as a real number) only when a and c have the same sign, so that ac > 0.
Arithmetic mean: 2b = a + c (the average). Geometric mean: b² = ac (the multiplicative middle). Three numbers form a geometric run iff the middle squared equals the product of the outer two.
Because every term is a₁·q to some power, the exponents simply add when you multiply terms. Take any two terms aₘ and aₙ:
aₘ · aₙ = (a₁·qm−1)(a₁·qn−1) = a₁²·qm+n−2.
That answer depends only on m + n. So whenever m + n = p + q, we get aₘ · aₙ = aₚ · a_q — terms whose indices add to the same total share the same product. In particular, pairs balanced around the ends satisfy
a₁ · aₙ = a₂ · aₙ₋₁ = a₃ · aₙ₋₂ = ⋯
This is the multiplicative twin of the arithmetic rule aₘ + aₙ = aₚ + a_q. In 2, 6, 18, 54 (so n = 4): a₁·a₄ = 2·54 = 108 and a₂·a₃ = 6·18 = 108 ✓ — equal, just as promised.
In a geometric sequence a₃ · a₇ = 20. Find a₅. Since 3 + 7 = 5 + 5, the property gives a₃·a₇ = a₅·a₅ = a₅², so a₅² = 20 and a₅ = ±√20 = ±2√5. The middle term is the geometric mean of the two outer ones — section 26.3.3 in disguise.
Write out the running total Sₙ = a₁ + a₁q + a₁q² + ⋯ + a₁qn−1. There is a beautiful trick to collapse it: multiply the whole thing by q and subtract. Multiplying shifts every term one slot to the right:
Subtracting the second line from the first, every middle term cancels in pairs, leaving only the ends:
Sₙ − qSₙ = a₁ − a₁qⁿ ⟹ (1 − q)Sₙ = a₁(1 − qⁿ).
Now divide by 1 − q — but only if you are allowed to. When q ≠ 1:
Sₙ = a₁(1 − qⁿ)1 − q (q ≠ 1).
Check it on 1, 2, 4, 8, 16 (a₁ = 1, q = 2, n = 5): S₅ = (1 − 2⁵)/(1 − 2) = (1 − 32)/(−1) = (−31)/(−1) = 31, and indeed 1+2+4+8+16 = 31 ✓.
If q = 1 the denominator 1 − q is 0 — the formula divides by zero and is forbidden. But q = 1 just means every term equals a₁, so the sum is trivial: Sₙ = n·a₁. Always check whether q = 1 before reaching for the fraction.
Here is what makes geometric sequences special. When |q| < 1, the power qⁿ shrinks toward 0 as n grows. Feed that into the sum formula and watch what happens to a₁qⁿ:
Sₙ = a₁(1 − qⁿ)1 − q → a₁(1 − 0)1 − q = a₁ / (1 − q) (as n → ∞).
So the running total creeps toward a finite number — it never overshoots, just edges closer and closer:
S = a₁1 − q only when |q| < 1.
This is Zeno's paradox made tame. Walk half the way to a wall, then half of what remains, then half again: 12 + 14 + 18 + ⋯ Here a₁ = ½ and q = ½, so S = (½)/(1 − ½) = (½)/(½) = 1. Infinitely many steps, but you arrive at exactly 1 — never a hair past it.
Fix a₁ = 1 and dial q in (−1, 1). The green dots are the running totals Sₙ; the green dashed line is the limit S = a₁/(1−q). The sums climb toward it but never touch.
If |q| ≥ 1 the terms do not shrink (they hold steady or grow), so the total runs off to infinity — there is no infinite sum. And q may be negative: when q = −½ the terms alternate sign (1, −½, ¼, −⅛, …) and the sum still converges, here to 1/(1−(−½)) = 2/3. The rule is about magnitude: |q| < 1.
A geometric sequence multiplies by a fixed ratio q; its term formula is a power of q, so it behaves like an exponential. The sum has a closed form for any finite n, and when the terms shrink (|q| < 1) even the infinite sum lands on a number.
| Arithmetic (26.2) | Geometric (26.3) | |
|---|---|---|
| step rule | add d = aₙ₊₁ − aₙ | multiply by q = aₙ₊₁ / aₙ |
| n-th term aₙ | a₁ + (n−1)d (linear) | a₁·qn−1 (exponential) |
| middle of a, b, c | 2b = a + c | b² = ac |
| sum Sₙ (n terms) | n(a₁+aₙ)/2 | a₁(1−qⁿ)/(1−q), q ≠ 1 |
| the trap | — | q = 1 ⟹ Sₙ = n·a₁ |
| infinite sum | never finite | a₁/(1−q) when |q| < 1 |
Next, 26.4 — Common methods for summing turns this "multiply, shift, and subtract" trick into one of four general tools for collapsing sums that are neither purely arithmetic nor purely geometric.
Ratios: 6÷3 = 2, 12÷6 = 2, 24÷12 = 2 — constant, so yes, q = 2. With a₁ = 3: aₙ = 3 · 2n−1.
a₅ = 5 · 3⁴ = 5 · 81 = 405.
b² = 4·9 = 36, so b = ±6. The runs 4, 6, 9 (q = 3/2) and 4, −6, 9 (q = −3/2) both work.
S₆ = 2(1 − 3⁶)/(1 − 3) = 2(1 − 729)/(−2) = 2(−728)/(−2) = 728. (Check: 2+6+18+54+162+486 = 728 ✓.)
Here a₁ = 1 and q = ⅓, with |q| < 1, so it converges: S = 1/(1 − ⅓) = 1/(⅔) = 3/2.
Here q = 1, so the formula divides by 1 − 1 = 0 — it does not apply. Use the q = 1 case: S₄ = 4 · 7 = 28.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops geometric sequences as exponential functions of the index, matching HSF-BF.A.2 (write arithmetic and geometric sequences with explicit and recursive forms) and HSF-LE.A.2 (construct geometric sequences given a description, terms, or a ratio). The closed forms for finite and infinite geometric series — including the convergence condition |q| < 1 — align with HSA-SSE.B.4 (derive and use the formula for the sum of a finite geometric series). Throughout, students practice attending to the q = 1 special case and to the magnitude condition on q, modeling the standards' emphasis on precision (MP6).