# How do you find three consecutive odd integers such that twice the sum of the first and third integers is 21 more than the second integer?

Oct 20, 2015

The three consecutive odd numbers are $5 , 7 , 9$

#### Explanation:

Let
$\textcolor{w h i t e}{\text{XXX}}$first odd number $= \textcolor{red}{2 n - 1}$
$\textcolor{w h i t e}{\text{XXX}}$second odd number $= \textcolor{b l u e}{2 n + 1}$
$\textcolor{w h i t e}{\text{XXX}}$third odd number $= \textcolor{b r o w n}{2 n + 3}$

We are told
$\textcolor{w h i t e}{\text{XXX}} 2 \left(\textcolor{red}{\left(2 n - 1\right)} + \textcolor{b r o w n}{\left(2 n + 3\right)}\right) = \textcolor{b l u e}{\left(2 n + 1\right)} + 21$

Simplifying
$\textcolor{w h i t e}{\text{XXX}} 8 n + 4 = 2 n + 22$

$\textcolor{w h i t e}{\text{XXX}} 6 n = 18$

$\textcolor{w h i t e}{\text{XXX}} n = 3$

So the three odd numbers are
$\textcolor{w h i t e}{\text{XXX}}$first $= 2 n - 1 = 2 \left(3\right) - 1 = 5$
$\textcolor{w h i t e}{\text{XXX}}$second $= 2 \left(3\right) + 1 = 7$
$\textcolor{w h i t e}{\text{XXX}}$third $= 2 \left(3\right) + 3 = 9$