arithmetic progression

Arithmetic progression of length $n$, initial term $a_1$ and common difference $d$ is the sequence $a_1,a_1+d,a_1+2d,\mathrm{\dots },a_1+(n-1)d$.

The sum of terms of an arithmetic progression can be computed using Gauss’s trick:

$S$$=(a_1+0)+(a_1+d)+\mathrm{\cdots }+(a_1+(n-2)d)+(a_1+(n-1)d)$$+\underset{¯}{S}$$\underset{¯}{=(a_1+(n-1)d)+(a_1+(n-2)d)+\mathrm{\cdots }+(a_1+d)+(a_1+0)}$$2S$$=(2a_1+(n-1)d)+(2a_1+(n-1)d)+\mathrm{\cdots }+(2a_1+(n-1)d)+(2a_1+(n-1)d).$

We just add the sum with itself written backwards, and the sum of each of the columns equals to $(2a_1+(n-1)d)$. The sum is then

$$S=\frac{(2a_1+(n-1)d)n}{2}.$$