If a series is arithmetic the sum of the first $n$ terms, denoted ${S}_{n}$, there are ways to find its sum without actually adding all of the terms.

To find the sum of the first $n$ terms of an arithmetic series use the formula, $n$
terms of an arithmetic sequence use the formula,

${S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}$,

where $n$
is the number of terms, ${a}_{1}$
is the first term and ${a}_{n}$
is the last term.

The series $3+6+9+12+\cdots +30$ can be expressed as sigma notation $\sum _{n=1}^{10}3n$. This expression is read as the sum of $3n$ as $n$ goes from $1$ to $10$

**Example 1:**

Find the sum of the first $20$ terms of the arithmetic series if ${a}_{1}=5$ and ${a}_{20}=62$.

$\begin{array}{l}{S}_{20}=\frac{20(5+62)}{2}\\ {S}_{20}=670\end{array}$

**Example 2:**

Find the sum of the first $40$
terms of the arithmetic sequence

$2,5,8,11,14,\cdots $

First find the $40$^{th} term:

$\begin{array}{l}{a}_{40}={a}_{1}+(n-1)d\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2+39\left(3\right)=119\end{array}$

Then find the sum:

$\begin{array}{l}{S}_{n}=\frac{n({a}_{1}+{a}_{n})}{2}\\ {S}_{40}=\frac{40(2+119)}{2}=2420\end{array}$

**Example 3:**

Find the sum:

$\sum _{k=1}^{50}(3k+2)$

First find ${a}_{1}$ and ${a}_{50}$:

$\begin{array}{l}{a}_{1}=3\left(1\right)+2=5\\ {a}_{20}=3\left(50\right)+2=152\end{array}$

Then find the sum:

$\begin{array}{l}{S}_{k}=\frac{k({a}_{1}+{a}_{k})}{2}\\ {S}_{50}=\frac{50(5+152)}{2}=3925\end{array}$