Formula for Arithmetic Series

The Simple Arithmetic Sequences

Let's say we have the simplest of arithmetic sequences.

$\{1, 2, 3, \cdot\cdot\cdot, n\}$

And what I want to think about is what is the sum of this sequence going to be? And the sum of a sequence, we already know we call a series as following:

$S_n = 1 + 2 + 3 + \cdot\cdot\cdot + n$

$S_n = n + (n-1) + (n-2) + \cdot\cdot\cdot + 1$

Now I'm going to add these two equations.

$2S_n = (n+1) + (n+1) + (n+1) + \cdot\cdot\cdot + (n+1)$

So how many of these $(n+1)$ do we have? Well we have n of them there were n of these terms in each of these equations. So, we can rewrite this thing as following:

$2S_n = n(n+1)$

$$ S_n = \frac{n(n+1)}{2} = n \cdot \frac{n+1}{2} = n \cdot \frac{a_n+a_1}{2} $$

$a_n$ is the nth term in our sequence, $a_1$ is the first term in our sequence.

General Arithmetic Sequences

Let's write an arithmetic sequence in general terms.

$\{a, a+d, a+2d,\cdot\cdot\cdot, a+(n-1)d\}$

d  could be a positive or a negative number, which we call our common difference.

So the arithmetic series is just the sum of an arithmetic sequence.

$S_n = a + (a+d) + (a+2d) + \cdot\cdot\cdot + [a+(n-1)d]$

I'm going to add this to itself, but I'm going to swap the order in which I write this sum.

$S_n = [a+(n-1)d] + [a+(n-2)d] + [a+(n-3)d] + \cdot\cdot\cdot + a$

Now let's add these two equations.

$2S_n = [2a+(n-1)d] + [2a+(n-1)d] + [2a+(n-1)d] + \cdot\cdot\cdot + [2a+(n-1)d]$

So you have this $[2a+(n-1)d]$ being added over and over again. And how many times are you doing that? Well, you had n pairs of terms when you were adding these two equations. In each of them you had n terms. So I can rewrite 2 times the sum as follows:

$$2S_n = n [2a+(n-1)d]$$

$$S_n = \frac{n [2a+(n-1)d]}{2}$$

Now, we've come up with a general formula, just a function of what our first term is, what our common difference is, and how many terms we're adding up. And so this is the generalized sum of an arithmetic sequence, which we call an arithmetic series.

But now, let's ask ourselves this question. This is hard to remember.

So let's rewrite the equation as following:

$$S_n = n\cdot\frac{a + [a+(n-1)d]}{2} =  n\cdot\frac{a_1 + a_n}{2}$$