# Sum of the First $n$ Terms of a Series

The sum of the terms of a sequence is called a series .

If a sequence is arithmetic or geometric there are formulas to find the sum of the first $n$ terms, denoted ${S}_{n}$ , without actually adding all of the terms.

(Note that a sequence can be neither arithmetic nor geometric, in which case you'll need to add using brute force, or some other strategy.)

## Sum of the Terms of an Arithmetic Sequence (Arithmetic Series)

To find the sum of the first $n$ terms of an arithmetic sequence use the formula,
${S}_{n}=\frac{n\left({a}_{1}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}{a}_{2}\right)}{2}$ ,
where $n$ is the number of terms, ${a}_{1}$ is the first term and ${a}_{n}$ is the last term.

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:

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

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:

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

## Sum of the Terms of a Geometric Sequence (Geometric Series)

To find the sum of the first $n$ terms of a geometric sequence use the formula,
${S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r},\text{\hspace{0.17em}}\text{\hspace{0.17em}}r\ne 1$ ,
where $n$ is the number of terms, ${a}_{1}$ is the first term and $r$ is the common ratio .

Example 4:

Find the sum of the first $8$ terms of the geometric series if ${a}_{1}=1$ and $r=2$ .

${S}_{8}=\frac{1\left(1-{2}^{8}\right)}{1-2}=255$

Example 5:

Find ${S}_{10}$ of the geometric series $24+12+6+\cdots$ .

First, find $r$

$r=\frac{{r}_{2}}{{r}_{1}}=\frac{12}{24}=\frac{1}{2}$

Now, find the sum:

${S}_{10}=\frac{24\left(1-{\left(\frac{1}{2}\right)}^{10}\right)}{1-\frac{1}{2}}=\frac{3069}{64}$

Example 6:

Evaluate.

$\underset{n=1}{\overset{10}{\sum }}3{\left(-2\right)}^{n-1}$

(You are finding ${S}_{10}$ for the series $3-6+12-24+\cdots$ , whose common ratio is $-2$ .)

$\begin{array}{l}{S}_{n}=\frac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}\\ {S}_{10}=\frac{3\left[1-{\left(-2\right)}^{10}\right]}{1-\left(-2\right)}=\frac{3\left(1-1024\right)}{3}=-1023\end{array}$