Geometric Sequences and Finite Series

A geometric sequence multiplies by a fixed common ratio r between consecutive terms, instead of adding a fixed amount like an arithmetic sequence does — scaling instead of stepping.

By the end you'll be able to find the nth term \(a_n = a_1\cdot r^{n-1}\) and the sum of the first n terms of a geometric sequence, and tell it apart from an arithmetic one.

Predict: with \(a_1 = 2\) and \(r = 3\), what is the 4th term? Set the sliders below and check.

Both sequences start at the same a₁. The geometric series multiplies by r each step; the arithmetic series (for comparison) adds d each step. Drag n to move the highlighted term and watch both readouts update. Hover or focus any dot for its exact value.

Term n = 3: geometric \(a_3 = a_1\cdot r^{2} = 1\cdot 4 = \) 4.00 — arithmetic \(a_3 = a_1 + 2d = 1 + 4 = \) 5.00

Geometric (×r) vs arithmetic (+d) — same starting term a₁, different growth rule
geometric: aₙ = a₁·r^(n−1) arithmetic: aₙ = a₁ + (n−1)d

A geometric sequence scales by the same factor r every step, producing exponential growth (\(|r|>1\)) or decay (\(|r|<1\)) — a fundamentally different shape than an arithmetic sequence's straight-line growth.

Intuitive

Where an arithmetic sequence adds the same amount each step, a geometric sequence scales by the same factor each step. Multiplying by r a total of \(n-1\) times (since the 1st term needs no multiplying) gives every later term, so the pattern compounds — small changes in r produce big changes far down the sequence, the same way compound interest outpaces simple interest.

Formal

With first term \(a_1\) and common ratio r, the closed nth-term formula is \(a_n = a_1\cdot r^{n-1}\). The sum of the first n terms comes from a clever cancellation: write \(S_n = a_1 + a_1 r + \cdots + a_1 r^{n-1}\), multiply by r, and subtract — nearly every term cancels, leaving

\(S_n = \dfrac{a_1(1 - r^n)}{1 - r}\), for \(r \neq 1\)

(If \(r=1\), every term equals \(a_1\), so \(S_n = n\cdot a_1\).) This finite-sum formula is the foundation for the infinite case, covered next.

Applied

A population doubling every year, a radioactive sample halving every half-life, money compounding at a fixed rate — all are geometric sequences. Once you know \(a_1\) and r, the nth-term formula tells you the value at any future step without listing every term in between.

Worked example

Find the 6th term and the sum of the first 6 terms of 5, 10, 20, 40, ...
Identify \(a_1 = 5\) and \(r = 10/5 = 2\).
\(a_6 = a_1\cdot r^{5} = 5\cdot 2^5 = 5\cdot 32 = \) 160.
\(S_6 = \dfrac{a_1(1-r^6)}{1-r} = \dfrac{5(1-2^6)}{1-2} = \dfrac{5(-63)}{-1} = \) 315.

Your turn

Same idea, a different sequence: 3, 12, 48, 192, ... Here \(a_1 = 3\) and \(r = 12/3 = 4\).
\(a_4 = a_1\cdot r^{3} = 3\cdot 4^3 = 3\cdot 64 = 192\).
Now finish the sum: \(S_4 = \dfrac{a_1(1-r^4)}{1-r} = \dfrac{3(1-256)}{1-4} = \) ____

Reveal the answer

\(S_4 = \dfrac{3(-255)}{-3} = \) 255. Set \(a_1 = 3\) as close as the slider allows and \(r\) near 4 on the demo above, then check \(a_4\) against the highlighted-term readout.

More info — why the sum formula works (the multiply-and-subtract trick)

Write out \(S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}\). Multiply every term by r: \(rS_n = a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1} + a_1 r^n\). Subtracting the first equation from the second cancels every middle term, leaving \(rS_n - S_n = a_1 r^n - a_1\), so \(S_n(r-1) = a_1(r^n - 1)\), which rearranges to \(S_n = a_1(1-r^n)/(1-r)\). See the Math is Fun link in Dive deeper for the full derivation with more examples.

Check your understanding

Question 1 of 4

A geometric sequence starts 4, 8, 16, 32, ... Find \(a_6\).

Question 2 of 4

Given consecutive terms of a geometric sequence, how do you find the common ratio r?

Question 3 of 4

A geometric sequence starts 3, -6, 12, -24, ... (so \(r = -2\)). Find \(a_5\).

Question 4 of 4

Sequence A: 5, 9, 13, 17, ... Sequence B: 5, 15, 45, 135, ... Which one is geometric, and what is its common ratio?

Recap

  • A geometric sequence multiplies by a constant common ratio r between terms; an arithmetic sequence instead adds a constant common difference d.
  • nth-term formula: \(a_n = a_1\cdot r^{n-1}\) — the exponent is \(n-1\), not n.
  • Finite-sum formula: \(S_n = a_1(1-r^n)/(1-r)\) for \(r \neq 1\) (and \(S_n = n\cdot a_1\) when \(r = 1\)).
  • Find r by DIVIDING consecutive terms, never by subtracting them.
  • A negative r alternates the sign of the terms; an even exponent flips it back positive.

Dive deeper

Sources

  • Geometric Sequences and Finite Series