# How many terms of the arithmetic sequence {1,3,5,7,...} will give a sum of 961?

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Alan P. Share
Dec 7, 2015

$31$ terms

#### Explanation:

For an arithmetic sequence with initial value $a$ and a difference between terms of $d$
the sum of the first $n$ terms is given by the formula
$\textcolor{w h i t e}{\text{XXX}} \Sigma = \frac{n}{2} \cdot \left(2 \textcolor{c y a n}{a} + \left(n - 1\right) \textcolor{g r e e n}{d}\right)$

For the given sequence
$\textcolor{w h i t e}{\text{XXX}} \textcolor{c y a n}{a = 1}$
and
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{d = 2}$

We are told that the required sum is $961$

So
$\textcolor{w h i t e}{\text{XXX}} 961 = \frac{n}{2} \cdot \left(2 \left(\textcolor{c y a n}{1}\right) + \left(n - 1\right) \left(\textcolor{g r e e n}{2}\right)\right)$

$\textcolor{w h i t e}{\text{XXX}} 961 = \frac{n}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} + \frac{n}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2}}}} \left(n - 1\right) \left(\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2}}}\right)$

$\textcolor{w h i t e}{\text{XXX}} 961 = n + {n}^{2} - n$

$\textcolor{w h i t e}{\text{XXX}} {n}^{2} = 961$

$\textcolor{w h i t e}{\text{XXX}} n = 31$

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