Solve the equation for $x$:
$1+4+7+10+.....+x=287$

Given that,

$1+4+7+10+....+x=287$

$a_n=a+\left(n-1\right)d$

$x=1+\left(n-1\right)\left(3\right)=3n-2$

Also, $S_n=\frac{n}{2}\left(a+l\right)$

$287=\frac{n}{2}\left(1+x\right)$

$287=\frac{n}{2}\left(1+3n-2\right)$

$574=n\left(3n-1\right)$

$=3n^2-n-574=0$

$=\left(3n+41\right)\left(n-14\right)=0$

$n=\frac{-41}{3},14$

We have $n=14$

Therefore, $x=3n-2=3\left(14\right)-2=40$