# Sum of the First $n$ Terms of an Arithmetic Series

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,
${S}_{n}=\frac{n\left({a}_{1}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{a}_{n}\right)}{2}$ ,
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 $\underset{n=1}{\overset{10}{\sum }}3n$ . 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 ${a}_{20}=62$ .

$\begin{array}{l}{S}_{20}=\frac{20\left(5\text{\hspace{0.17em}}+\text{\hspace{0.17em}}62\right)}{2}\\ {S}_{20}=670\end{array}$

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:

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

Then find the sum:

$\begin{array}{l}{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\\ {S}_{40}=\frac{40\left(2\text{\hspace{0.17em}}+\text{\hspace{0.17em}}119\right)}{2}=2420\end{array}$

Example 3:

Find the sum:

$\underset{k=1}{\overset{50}{\sum }}\left(3k+2\right)$

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:

$\begin{array}{l}{S}_{k}=\frac{k\left({a}_{1}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{a}_{k}\right)}{2}\\ {S}_{50}=\frac{50\left(5\text{\hspace{0.17em}}+\text{\hspace{0.17em}}152\right)}{2}=3925\end{array}$