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.