This post is about quickly adding a sequential set of numbers from an arithmetic sequence. An arithmetic sequence is a list of numbers that have the same difference between each two adjacent numbers.

For example: 2, 4, 6, 8, 10, … is an arithmetic sequence with a difference of 2 between each two numbers. What if you needed to add the first 20 of these numbers or the next 20 numbers starting with any number in the sequence? Even if you don’t need to do this, the following is very interesting and surprising.

Let’s start with this arithmetic sequence: 7, 14, 21, …, 63, 70. You can see that the difference between the numbers is 7. If you write this sequence down, add the first and last number: 7 + 70 = 77. Now add the second number and the next to the last number: 14 + 63 = 77. What?! Will I get the same number if I keep doing this? It turns out, that you will:

There are 5 pairs of 77’s so 5 × 77 = 385 and this is the same answer you would get if you added all 10 numbers manually.

It turns out that this trick will work no matter what the difference between each number, how many numbers that are to be added, or where you start adding in the sequence. The only condition is that you are using an arithmetic sequence and that the numbers added are sequential, that is you can’t skip numbers.

So a generic sequence addition looks like this:

\[{a}_{1}\hspace{0.33em}{+}\hspace{0.33em}{a}_{2}\hspace{0.33em}{+}\hspace{0.33em}{a}_{3}\hspace{0.33em}{+}\hspace{0.33em}\cdots\hspace{0.33em}{+}\hspace{0.33em}{a}_{{n}{-}{1}}\hspace{0.33em}{+}\hspace{0.33em}{a}_{n}

\]

The subscripts just keep track of the order of the terms. The last term, *a*_{n}, means that there are *n* numbers to be added. In our example, *n* was 10 and *a*_{n} was 70.

So as in our example, we added the first and last term, then multiplied that by half the numbers to be added, that is *n*/2. The formula for this trick using the generic sequence is:

{\mathrm{Sum}}\hspace{0.33em}{=}\hspace{0.33em}\frac{n}{2}\hspace{0.33em}\times\hspace{0.33em}{(}{a}_{1}\hspace{0.33em}{+}\hspace{0.33em}{a}_{n}{)}

\]

This works even if *n* is odd. What I find fascinating about this formula is that it doesn’t include the difference between each number. It doesn’t matter what the difference between each number is. This formula just needs to know the first and last number and how many numbers are to be added. Isn’t math strange and wonderful!