## Precalculus (6th Edition) Blitzer

The sum, ${{S}_{n}}$, of the first $n$ terms of the sequence described in Exercise 1 is given by the formula ${{S}_{n}}$ = $\frac{n}{2}({{a}_{1}}+{{a}_{n}})$, where ${{a}_{1}}$ is the first term and ${{a}_{n}}$ is the nth term.
To find the sum of the first n terms, we need to have the first term of the sequence, the common difference and the number of terms of which sum is to be determined. The sum is denoted by ${{S}_{n}}$, the first term by ${{a}_{n}}$, the common difference by $d$, and the number of terms by $n$. Thus, the formula for determining the sum of the first n terms is given by, \begin{align} & {{S}_{n}}=\frac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right) \\ & {{S}_{n}}=\frac{n}{2}\left\{ 2{{a}_{1}}+\left( n-1 \right)d \right\} \\ \end{align}