Arithmetic sequence problem $\frac{x+4}{x-3},\frac{x+6}{2},\frac{4}{x-2}$ Choose such x that the following
$$\frac{x+4}{x-3},\frac{x+6}{2},\frac{4}{x-2}$$
forms finite arithmetical sequence.
If I use the equation: $$2a_2=a_3+a_1$$
I always get the wrong answer.

Let 
$$
a_1=\frac{x+4}{x-3}, \quad a_2=\frac{x+6}{2},\quad a_3=\frac{4}{x-2}
$$
If $ a_1 $, $ a_2 $ and $ a_3 $ are in arithmetic progression the possible values for $ x $ are
$$
4,\quad \frac{-4+\sqrt{72}}{2} \quad\mbox{and}\quad \frac{-4-\sqrt{72}}{2}
$$
We have $2a_2={a_3+a_1}$, that is,
$$
{x+6}=\frac{x+4}{x-3}+\frac{4}{x-2}
$$ 
\begin{align}
(x+6)(x-3)(x-2)=(x+4)(x-2)+4(x-3)
\\
\\
[(x+6)(x-3)-(x+4)](x-2)-4(x-3)=0
\\
\\
[x^2+2x-22](x-2)-4x+12=0
\\
\\
x^3-30x+56=0
\end{align}
Note that $x=4$ is a integer  solution this equation.
$$
x^3-30x+56=(x-4)(x^2+4x-14)=0
$$
The other two roots are
$$
\frac{-4+\sqrt{72}}{2} \quad \frac{-4-\sqrt{72}}{2}
$$
