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Arithmetic Series

An arithmetic series is a series whose related sequence is arithmetic.  It results from adding the terms of an arithmetic sequence .

Example 1:

Finite arithmetic sequence: $5,10,15,20,25,...,200$

Related finite arithmetic series: $5+10+15+20+25+...+200$

Written in sigma notation: $\underset{k=1}{\overset{40}{\sum }}5k$

Example 2:

Infinite arithmetic sequence: $3,7,11,15,19,...$

Related infinite arithmetic series: $3+7+11+15+19+...$

Written in sigma notation: $\underset{n=1}{\overset{\infty }{\sum }}\left(4n-1\right)$

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

Example 3:

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

$\begin{array}{l}{S}_{10}=\frac{20\left(5+62\right)}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=670\end{array}$

Example 4:

Find the sum of the first $40$ terms of the arithmetic series
$2+5+8+11+...$ .

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

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

Then find the sum:

$\begin{array}{l}{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\\ {S}_{10}=\frac{40\left(2+119\right)}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2420\end{array}$

Example 5:

Find the sum:

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

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

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

Then find the sum:

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

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