Editing Arithmetic progression (section)

=== Derivation ===
[[File:Animated proof for the formula giving the sum of the first integers 1+2+...+n.gif|thumb|Animated proof for the formula giving the sum of the first integers 1+2+...+n.]]
To derive the above formula, begin by expressing the arithmetic series in two different ways:
:<math> S_n=a+a_2+a_3+\dots+a_{(n-1)} +a_n</math>
:<math> S_n=a+(a+d)+(a+2d)+\dots+(a+(n-2)d)+(a+(n-1)d). </math>
Rewriting the terms in reverse order:
:<math> S_n=(a+(n-1)d)+(a+(n-2)d)+\dots+(a+2d)+(a+d)+a.</math>

Adding the corresponding terms of both sides of the two equations and halving both sides:
:<math> S_n=\frac{n}{2}[2a + (n-1)d].</math>
This formula can be simplified as:
:<math>\begin{align}
S_n &=\frac{n}{2}[a + a + (n-1)d].\\
&=\frac{n}{2}(a+a_n).\\
&=\frac{n}{2}(\text{initial term}+\text{last term}).
\end{align}</math>
Furthermore, the mean value of the series can be calculated via: <math>S_n / n</math>:
:<math> \overline{a} =\frac{a_1 + a_n}{2}.</math>

The formula is essentially the same as the formula for the mean of a [[discrete uniform distribution]], interpreting the arithmetic progression as a set of equally probable outcomes.