Question

Find four consecutive even integers so that the sum of the first two added to twice the sum of the last two is equal to $742$.

Answer 1

Here let the four even integers be $x$, $x+2$, $x+4$ and $x+6$ then the equation as per the question will be,
$\begin{array}{cc}x+x+2+2\left(x+4+x+6\right)& =742\\ 2x+2+2\left(2x+10\right)& =742\\ 2x+2+4x+20& =742\\ 6x+22& =742\\ 6x& =720\\ x& =120\end{array}$

Now, the four consecutive even integers are
$x=120$
$x+2=122$
$x+4=124$
$x+6=126$