Contests Love Series

In arguably every contest you do, there will be some mention/use-case involving series or a sequence. This module will cover some of the fundamentals of series, (specifically, arithmetic series), tread carefully!

Sequence: An ordered list of values, e.g. $1, 7, 13, 19$

Series: The sum of the numbers in a sequence, e.g. $1 + 7 + 13 + 19 = 40$

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, instead, 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

To define a sequence, we usually use notation such as $a_1, a_2, ..., a_n$ where $n$ is the number of terms in the sequence. However, you may use any letter you wish to use.

An arithmetic sequence, as discussed above, depends on two things:

1. A given term of the sequence, $a_k$
2. The common difference, $d$

For instance, suppose you are given $a_1$, then, $a_5 = a_1 + 4d$. Generally:

Given $k \le n$, $a_n = a_k + (n-k)d$

The reason this works is that moving from term $k$ to term $n$ means adding the common difference exactly $n - k$ times. If you know any single term and the common difference, you can reach every other term in the sequence, so you never actually need to know $a_1$ to describe the whole thing. This flexibility is useful on contests where you are handed a term in the middle rather than the start.

For instance, $2, 4, 6, 8$ is an arithmetic sequence with common difference $2$ and $109, 99, 89, 79$ is an arithmetic sequence with common difference $-10$; however, $7, 67, 1, 90$ and $1, 1010, 3, 4, \ldots$ are not arithmetic sequences, as the difference between consecutive terms varies.

Formally, the sequence $a_1, a_2, \ldots , a_n$ is an arithmetic progression if and only if $a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}$.

Core Skills

Gauss' Genius

In the late 1700s, a schoolmaster named J.G. Büttner wanted to keep his primary school class quiet and busy for an hour. He gave his students what he thought was a tedious arithmetic chore: add all the whole numbers from $1$ to $100$. While the other children immediately began writing out $1 + 2 = 3$, $3 + 3 = 6$, $6 + 4 = 10$, and tracking long columns of numbers, a young Gauss sat quietly for a few moments. He then walked up to the teacher's desk and threw down his slate with the exact answer: $5,050$

How did he do this? By utilizing a core property that arithemtic sequences have. Specifically, for an arithmetic sequence: $a_1, a_2, ..., a_n$, the values $a_n - a_1, a_{n-1} - a_2, ...$ all remain constant, as you go up in the sequence by one and go down in the other, the changes cancel each other out!

The trick is to think in pairs rather than in single terms. Gauss imagined the numbers $1$ through $100$ written forwards, then the same numbers written backwards underneath: $1$ pairs with $100$, $2$ pairs with $99$, $3$ pairs with $98$, and so on. Every one of these pairs sums to $101$, and there are $100$ of them, giving $100 \times 101 = 10100$. Since each number got counted twice, the real answer is half of that, or $5050$. The strength of the idea is that it turns a long, error-prone addition into a single multiplication.

We can use this fact to motivate the formula for the sum of an arithmetic series:

$S = \dfrac{(a_1 + a_n) \cdot n}{2}$

That is, the sum of a series is the sum of the first and last terms, multiplied by the number of pairs, $n/2$.

The Sum Formula

The sum of the first $n$ terms is equal to:

The second form is handy when you are not told the last term directly, since you can substitute $a_n = a_1 + (n-1)d$ and work entirely from $a_1$, $d$, and $n$. Both forms describe the same quantity, so use whatever matches the information the problem hands you.

Proof. We want to show that $S_n = \frac{n}{2}(a_1 + a_n)$. Start by writing the sum out in full, then write the same sum a second time but with the terms reversed:

Now add the two lines together, lining up the columns. In the first column we get $a_1 + a_n$. In the second column we get $(a_1 + d) + (a_n - d) = a_1 + a_n$, since the $+d$ and $-d$ cancel. The same cancellation happens in every column, so each of the $n$ columns sums to exactly $a_1 + a_n$. Adding all of them gives

Dividing both sides by $2$ yields

which is what we wanted. To get the second form, substitute $a_n = a_1 + (n-1)d$:

This is also known as Gauss's trick. Notice that it is exactly the pairing idea from the story above, just written algebraically and made to work for any arithmetic sequence rather than only $1$ through $100$.

Finding the Number of Terms

Hence given $a_1$, $d$, and the last term $a_n$:

Always solve for $n$ before applying the sum formula, as this can be a common source for errors off by one term. The intuition for the $+1$ is that $\frac{a_n - a_1}{d}$ counts the number of steps between the first and last terms, but a sequence with one step has two terms, a sequence with two steps has three terms, and so on.

Arithmetic Mean

The average of an arithmetic sequence always equals the average of the first and last terms:

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. The reason the average sits at the midpoint is that the terms are spread out evenly, so whatever lies above the center is balanced by an equal amount below it. This is also why the sum formula can be read as "average term times number of terms," which is the same statement as $S_n = n \cdot \bar{a}$.

A useful equation is the sum of the first $n$ positive integers,

This is the arithmetic sequence $a_1 = 1$, $d = 1$ with sum formula applied. It is worth memorizing on its own, since it appears constantly in counting problems, in the number of edges of a complete graph, and inside larger algebra manipulations.

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. The safe way to do these problems is to find the smallest valid value at or above $a$, find the largest valid value at or below $b$, and then count the terms between them with the $n$ formula rather than estimating.

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 $S_n = \frac{25}{2}(3 + 99) = 25 \cdot 51 = 1275$.

Worked Examples

2004 AMC 12B · Problem 8 A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?

The number of cans per row, read from the top down, is $1, 3, 5, 7, \ldots$ This is an arithmetic sequence with $a_1 = 1$ and $d = 2$, so the total number of cans in $n$ rows is the sum of the first $n$ terms. We are told this sum is $100$, so we set up the sum formula:

So the number of rows satisfies $n^2 = 100$, giving $n = \boxed{10}$.

There is a slicker way to see the same thing. The cans per row are the first $n$ odd numbers, and the sum of the first $n$ odd numbers is always $n^2$ (try it: $1 = 1$, $1+3 = 4$, $1+3+5 = 9$). Setting $n^2 = 100$ again gives $n = 10$. The sum of the first $n$ odd numbers is indeed $n^2$, worth remembering!

2006 AMC 12A · Problem 12 A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?

Be careful!. Do not overcount where rings overlap. A nice way to set this up is to add the full outside diameters first, then subtract the overlaps. The outside diameters run $20, 19, 18, \ldots, 3$, which is an arithmetic sequence with $a_1 = 20$, $d = -1$, and last term $3$. The number of rings is

The sum of these outside diameters is

But stacking the rings creates an overlap at each link. Since each ring is $1$ cm thick, two ring thicknesses ($2$ cm) of vertical space are shared wherever one ring hangs inside the next. There are $17$ such links between the $18$ rings, so we subtract $2 \cdot 17 = 34$:

Practice Problems