Unit 9: Sequences, Series, and Combinatorics · ~20 min
Objective
Find any term of a geometric sequence, and evaluate the sum of a finite geometric series and, when it converges, an infinite one.
Where an arithmetic sequence adds the same amount d each step, a geometric sequence multiplies by the same amount r each step — the common ratio. This is the discrete version of the exponential functions from Unit 3: just as f(x) = a·\(b^{x}\) grows or decays by a constant factor for every unit increase in x, a geometric sequence grows or decays by the constant factor r for every step in n.
Explicit formula
\(a_{n}\) = \(a_{1}\) · \(r^{n-1}\)
The recursive form is \(a_{n}\) = r · \(a_{n-1}\), with \(a_{1}\) given to start it. The sign and size of r determine the sequence's behavior: |r| > 1 means the terms grow without bound, |r| < 1 means they shrink toward 0, and a negative r makes the terms alternate sign.
Finite geometric series
To sum the first n terms, multiply the sum \(S_{n}\) by r and subtract from the original sum — every term except the first and the r·\(a_{n}\) term cancels, leaving a formula that avoids adding n terms by hand:
If |r| < 1, each new term is smaller than the last, and as n → ∞ the term \(r^{n}\) shrinks toward 0. Substituting that limit into \(S_{n}\) gives a finite sum even though there are infinitely many terms — the series converges. If |r| ≥ 1, the terms don't shrink and the sum grows without bound — the series diverges, and has no sum.
Infinite geometric series (|r| < 1)
S = \(\frac{a_{1}}{1 - r}\)
\(a_{n}\) = 20(½\()^{n-1}\) — terms shrink toward 0 as n grows since |r| < 1
Worked Example 1 · Finding a term
ProblemA geometric sequence has \(a_{1}\) = 5 and common ratio r = 3. Find \(a_{6}\).