Arithmetic sequences grow by a fixed difference. They show up in many counting, number theory, and series problems. At the AMC level, the challenge is rarely about just plugging into a formula — it's about setting up the right sequence.
Beyond the basics, arithmetic sequences connect to divisibility (multiples of k form an arithmetic sequence), counting (how many integers in a range to satisfy a condition), and algebra (systems involving sequences).
Key Ideas
an=a1+(n−1)d.
Sn=2n(a1+an).
Pairing terms from the ends makes sums fast.
Core Skills
The General Term
For an arithmetic sequence with first term a1 and common difference d, we have the formula
an=a1+(n−1)d.
d>0: increasing sequence.
d<0: decreasing sequence.
d=0: constant sequence (still arithmetic).
The common difference is always d=ak+1−ak for any consecutive pair.
The Sum Formula
The sum of the first n terms is equal to:
Sn=2n(a1+an)=2n(2a1+(n−1)d).
Proof: Write the sum forwards and backwards and add, hence obtaining
S_n &= a_1 + (a_1+d) + \cdots + a_n \\ S_n &= a_n + (a_n-d) + \cdots + a_1 \end{aligned}$$ Adding gives $2S_n = n(a_1 + a_n)$, so $S_n = \frac{n}{2}(a_1 + a_n)$. This is also known as Gauss's trick. ### Finding the Number of Terms Hence given $a_1$, $d$, and the last term $a_n$: $$n = \frac{a_n - a_1}{d} + 1.$$ Always solve for $n$ before applying the sum formula, as this can be a common source for errors off by one term. ### Arithmetic Mean The average of an arithmetic sequence always equals the average of the first and last terms: $$\bar{a} = \frac{a_1 + a_n}{2} = a_{\text{middle}}.$$ For an odd number of terms, the middle term equals the mean. This makes the sum $S_n = n \cdot \bar{a}$ convenient to find when $n$ is known. A useful equation is the sum of the first $n$ positive integers, $$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}.$$ This is the arithmetic sequence $a_1 = 1$, $d = 1$ with sum formula applied. ### Multiples in a Range The number of multiples of $k$ in $\{1, 2, \ldots, N\}$ is $\lfloor N/k \rfloor$. Therefore, the multiples themselves form an arithmetic sequence $k, 2k, 3k, \ldots$ with $d = k$. More generally, the number of integers in $\{a, a+1, \ldots, b\}$ that are $\equiv r \pmod{k}$ is approximately $(b-a)/k$, can be found by finding the first and last terms and applying the $n$ formula. ## Worked Example Find the sum $3 + 7 + 11 + \cdots + 99$. This is an arithmetic sequence with $a_1 = 3$, $d = 4$, and $a_n = 99$. The number of terms is $n = \frac{99-3}{4} + 1 = 25$. So $S_n = \frac{25}{2}(3 + 99) = 25 \cdot 51 = 1275$. ## More Examples ### Example 1: Find a Term If $a_1=12$ and $d=-3$, find $a_6$. $a_6 = 12 + 5(-3) = -3$. ### Example 2: Find $n$ How many terms are in $5,9,13,\ldots,77$? $n = (77-5)/4 + 1 = 19$. ### Example 3: Pairing Compute $2 + 5 + 8 + \cdots + 62$ by pairing. There are $n=(62-2)/3+1=21$ terms, and each pair sums to $64$, so $S_n = 21\cdot 32 = 672$. ## Strategy Checklist - First identify $a_1$, $d$, and either $n$ or $a_n$. - Solve for $n$ before summing. - Pair terms when the list is long. - Check that the last term fits the sequence. - Check both $d > 0$ and $d < 0$, as both give valid sequences unless the problem specifies if it is increasing or decreasing. ## Common Pitfalls - Using $n-1$ incorrectly when solving for $n$. - Forgetting to check that the last term belongs to the sequence. - Treating a decreasing sequence as increasing and miscounting terms. ## Practice Problems <Problems problems="practice" />
