MISSION · Algebra
Arithmetic and Geometric Sequences
Two kinds of pattern, and the formulas that skip the counting
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The idea in one sentence
An arithmetic sequence adds the same amount each time; a geometric one multiplies by the same amount — and once you know which, you can reach the hundredth term without writing the first ninety-nine.
Telling them apart
Look at the gaps.
\(3, 7, 11, 15, 19, \ldots\) — each term is 4 more than the last. Arithmetic, with a common difference of \(d = 4\).
\(3, 6, 12, 24, 48, \ldots\) — each term is double the last. Geometric, with a common ratio of \(r = 2\).
If subtracting consecutive terms gives a constant, it’s arithmetic. If dividing them gives a constant, it’s geometric. If neither, it’s something else — the squares \(1, 4, 9, 16\) are neither, and so is the Fibonacci sequence, where each term is the sum of the two before it.
The nth-term formulas
\(\text{arithmetic:} \quad a_n = a_1 + (n - 1)d\) \(\text{geometric:} \quad a_n = a_1 \cdot r^{\,n-1}\)
Both have the same shape, and the same reason behind it. To reach the \(n\)th term from the first, you take \(n - 1\) steps — not \(n\). That off-by-one is where nearly every mistake in this topic comes from, and the fix is to check with a term you already know.
For \(3, 7, 11, \ldots\): \(a_n = 3 + 4(n - 1) = 4n - 1\). Test it on the third term: \(4(3) - 1 = 11\). Correct, so the hundredth term is \(4(100) - 1 = 399\), found without listing anything.
What the formulas are worth
The point isn’t the fifth term — you can count to that. It’s that the rule describes the whole infinite sequence in one line, so any term is one substitution away, and questions like “which term is the first to pass 1000?” become equations rather than searches.
Where they go afterwards
Arithmetic sequences are linear growth: plot them and the points sit on a straight line, with \(d\) as the slope. Geometric sequences are exponential growth, curving upward when \(r > 1\) and decaying towards zero when \(0 < r < 1\).
A series is what you get when you add a sequence’s terms instead of listing them, and one result there is worth meeting early: a geometric series with \(|r| < 1\) adds up to a finite total even with infinitely many terms. \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\) never exceeds 2 — the pieces shrink fast enough for the infinite sum to converge.
Common mix-ups
- Using \(n\) where the formula wants \(n - 1\). Count the steps, not the terms.
- Assuming any pattern is one of these two. Check the differences and the ratios before committing.
- Mixing up difference and ratio. Subtract for arithmetic, divide for geometric.
- Starting at \(a_0\) when the formula assumes \(a_1\). Be consistent about where the counting begins.
Check yourself
Find the 20th term of \(5, 8, 11, 14, \ldots\), then identify the type of \(80, 40, 20, 10, \ldots\) and give its rule.
The first is arithmetic with \(d = 3\), so \(a_{20} = 5 + 3(19) = 62\). The second is geometric with \(r = \frac{1}{2}\), so \(a_n = 80 \cdot (0.5)^{\,n-1}\).
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