Arithmetic sequence problem $\frac{x+4}{x-3},\frac{x+6}{2},\frac{4}{x-2}$

Question

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.

Answer 1

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 $$ \frac{-4+\sqrt{72}}{2} \quad \frac{-4-\sqrt{72}}{2} $$