Arithmetic progression of length $n$ , initial term $a_1$ and common difference $d$ is the sequence .

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

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 \begin{equation*} S=\frac{(2a_1+(n-1)d)n}{2}. \end{equation*}