The sum of an arithmetic series

Add the terms of an arithmetic sequence — a sequence that goes up by the same amount each time — and you get

where $a$ is the first term, $l$ the last, $d$ the common difference and $n$ the number of terms.

The proof, and the six-year-old who found it

There is a story — probably not true, and told anyway — that a teacher set a class the job of adding the numbers from $1$ to $100$ to keep them quiet. A pupil wrote the answer down almost at once. Here is what he did, generalised.

Write the sum down twice, the second time backwards.

Add the two lines, term by term.

The first terms give $a + l$. The second terms give $(a + d) + (l - d) = a + l$. The third terms give $(a + 2d) + (l - 2d) = a + l$ as well.

Every pair adds to the same thing, $a + l$, because the $d$ that one term gained the other term has lost. There are $n$ pairs, so adding the two lines gives

and therefore

Turning the last term into the common difference. If you are not told $l$ but are told $d$ instead, use $l = a + (n-1)d$, which is the rule for the $n$th term:

which is the second form.

Check it on the story

For $1 + 2 + \dots + 100$: $a = 1$, $l = 100$, $n = 100$, so $S = \tfrac{1}{2} \times 100 \times 101 = 5050$.

Check your understanding

1. Find the sum of the first $20$ terms of $3, 7, 11, \dots$.
2. Why must the two written-out sums be the same length?
3. A series has $a = 5$, $d = 4$ and $n = 30$. Which form of the rule would you use, and why?