SEQUENCES & SERIES

Overview

A sequence is an ordered list of numbers following a rule; a series is the sum of the terms of a sequence. On the AMC10, sequence and series problems appear at almost every difficulty level — from "find the 10th term of 3, 7, 11, …" to "find a parameter that forces a recursive sequence into a given pattern." Learning to recognize which type of sequence you're looking at is the first and most important step.

The two workhorses are arithmetic sequences (each term is obtained by adding a fixed number $d$, the common difference) and geometric sequences (each term is obtained by multiplying by a fixed number $r$, the common ratio). Most AMC problems either ask for a specific term, a partial sum, an infinite sum, or a hidden parameter — all of which follow from two or three formulas. Speed matters, so internalize those formulas deeply enough to apply them without hesitation.

Beyond the standard types, the AMC10 loves telescoping sums and recursive sequences. Telescoping exploits cancellation when each term can be written as a difference $f(k) - f(k+1)$. Recursive sequences reward pattern-finding: compute the first several terms and look for periodicity or a closed form before attempting algebra.

Key facts

• Arithmetic sequence — $n$th term: $a_n = a_1 + (n-1)d$ — use to find any term directly.
• Arithmetic series — partial sum: $S_n = \dfrac{n}{2}(a_1 + a_n) = \dfrac{n}{2}(2a_1 + (n-1)d)$ — sum the first $n$ terms without adding them one by one.
• Geometric sequence — $n$th term: $a_n = a_1 \cdot r^{n-1}$ — multiply the first term by the ratio raised to the $(n-1)$ power.
• Geometric series — partial sum: $S_n = a_1 \cdot \dfrac{1 - r^n}{1 - r}$ for $r \neq 1$ — use when $r$ is a simple fraction or integer.
• Infinite geometric series: $S = \dfrac{a_1}{1 - r}$, valid only when $|r| < 1$ — the series converges to a finite value.
• Telescoping: if $f(k) - f(k+1)$ can be factored out of each term, $\displaystyle\sum_{k=1}^{n} (f(k) - f(k+1)) = f(1) - f(n+1)$ — the middle terms cancel.
• Arithmetic mean: the middle term of three consecutive terms of an arithmetic sequence equals their average. For terms $a$, $b$, $c$ in arithmetic progression: $b - a = c - b$, i.e., $2b = a + c$.
• Geometric mean: for three terms $a$, $b$, $c$ in geometric progression: $b^2 = ac$.
• Recursive to closed form: compute several terms, look for a pattern, then verify with the recurrence. Common patterns include linear growth ($a_n = an + b$), exponential growth ($a_n = A \cdot r^n + C$), and periodicity.

Worked example 1

The first term of an arithmetic sequence is $5$ and the common difference is $3$. What is the sum of the first $10$ terms?

We use the partial-sum formula. With $a_1 = 5$, $d = 3$, and $n = 10$:

$S_{10} = \frac{10}{2}\bigl(2(5) + (10-1)(3)\bigr) = 5(10 + 27) = 5 \cdot 37 = 185.$

Why does this formula work? Write the sum forwards and backwards, then add:

$S = 5 + 8 + \cdots + 32, \qquad S = 32 + 29 + \cdots + 5.$

Each pair adds to $5 + 32 = 37$, and there are $10$ pairs, so $2S = 370$, giving $S = 185$. The general formula is just this "pair-up" idea made algebraic.

Worked example 2

The fractions $\dfrac{1}{1 \cdot 2},\ \dfrac{1}{2 \cdot 3},\ \dfrac{1}{3 \cdot 4},\ \ldots,\ \dfrac{1}{99 \cdot 100}$ are added together. What is their sum?

Each fraction telescopes via partial fractions:

$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.$

So the sum becomes:

$\sum_{k=1}^{99}\left(\frac{1}{k} - \frac{1}{k+1}\right) = \frac{1}{1} - \frac{1}{100} = \frac{99}{100}.$

Everything between the first $\dfrac{1}{1}$ and the last $-\dfrac{1}{100}$ cancels in pairs — that is the "telescope" collapsing. The trick is recognizing the partial-fraction split; once you see it, the sum writes itself.

Common traps

• Off-by-one in the $n$th term. The $n$th term is $a_1 + (n-1)d$, not $a_1 + nd$. The first term is $n = 1$, so there are only $n - 1$ steps of size $d$.
• Forgetting $|r| < 1$ for the infinite series. The formula $\dfrac{a_1}{1-r}$ only applies when the series converges. If $|r| \geq 1$, the sum is infinite (or undefined).
• Confusing ratio and difference. Checking whether a sequence is arithmetic or geometric is step one. If the differences of consecutive terms are equal, it's arithmetic; if the ratios are equal, it's geometric.
• Misidentifying the first term in a partial sum. When a problem gives you the 3rd and 7th terms, convert to $a_1$ first before plugging into the sum formula.
• Stopping the telescope too early. The telescoping sum collapses to $f(1) - f(n+1)$, not $f(1) - f(n)$. Use $f(n+1)$ at the last step.