You’ve got the formulas in the book, of course.

$u_n = a + (n-1)d$ $S_n = \frac n2 \left(a + L\right) = \frac n2 \left(2a + (n-1)d\right)$

This is somewhere the book and I have a serious disagreement: as a mathematical document, it ought to define its terms.

$a$ is the first term of the progression. $d$ is the common difference. $n$ is the number of the term (or sum) you’re looking for. $L$ is the last (or $n$th) term of the sequence. $u_n$ is the $n$th term of the progression. $S_n$ is the sum of the first $n$ terms.

Right, that’s better. But where do these come from? And is there another way?

### Where the $u_n$ formula comes from

It’s simple enough: the $n$th term is $(n-1)$ ‘steps’ above the first term, and each step has size $d$. So that’s hardly worth bothering with.

However, you can also think of an arithmetic progression as a straight line: if you’re told a specific term (for instance, the first term is 10), you can treat that as a point on the line: it would be $(1,10)$. If you’re told the common difference, you can treat that as the gradient. After that, the equation of the line should drop out - and be identical to the equation you had to begin with.

### How about the sum?

Well, the sum is trickier. The standard proof is to write the sum down in two different orders:

$\begin{array}{cccccccc} S_n = &a & + a+d & + a + 2d + & … & + a + (n-3)d & + a +(n-2)d & + a + (n-1)d \\ S_n = & a + (n-1)d & + a + (n-2)d & + a + (n-3)d & … & + a + 2d & + a + d & + a \end {array}$

Then add them up:

$2S_n = [2a + (n-1)d] + [2a + (n-1)d] + [2a + (n-1)d] + … + [2a + (n-1)d] + [2a + (n-1)d] + [2a + (n-1)d]$ … a total of $n$ identical terms on the right.

$2S_n = n(2a + (n-1)d)$, so $S_n = \frac n2 (2a + (n-1)d)$ - which is the same as $\frac n2 (a + L)$.

Another method is to think of the arithmetic series as a bar graph (strictly, a histogram). If you copied the graph and spun it around 180 degrees, it’d fit nicely on top of the histogram you’ve got, making a rectangle. The height of the rectangle would be $(a+L)$ and the width would be $n$, meaning $2S_n = n(a+L)$, like before.

It also, less than intuitively, is an integral: it’s $\int_{\frac12}^{n + \frac12} a + (x-1)d \, \d x$. Don’t believe me? Try it yourself. The limits seem a little bit odd, but they make sense in terms of the bar chart: $(1,a)$ is strictly in the middle of a bar, which would have the same area as the trapezium under the line starting half a unit lower and ending half a unit higher.