Example: Add up the numbers from $1$ to $100$.
There is a nice way to do this by writing the sum in two ways, adding them up and dividing by $2$.
$$\begin{array}{llllll}
1&+2&+3&+4&+\cdots &+100\\
100&+99&+98&+97&+\cdots &+ 1\\\hline
101&+101&+101&+101&+\cdots &+ 101
\end{array}$$
Noting that $101$ occurs $100$ times these add up to $100\times 101$.
Hence the sum
$$1+2+3+4+\cdots +100=\displaystyle {100\times 101 \over 2}$$
Did you notice that the first and last terms of the series add up to $101$. This is how the general case works too.
Sum of an Arithmetic Series
The sum of the arithmetic series with $n$ terms
$$A+ (A+d) + (A+2d)+\cdots + L$$
is
$$S_n={n(A+L)\over 2}$$
where
$A=$ the first term
$L=$ the last term
$n=$ the number of terms in the series
$L=$ the last term
For the example above,
$A=1$, $L=100$ and $n=100$ (and $d=1$ but we didn't need it here).
$A=1$, $L=100$ and $n=100$ (and $d=1$ but we didn't need it here).
