Geometric sequence

$\small{\color{goldenrod}{a,b,c~in~GP\Rightarrow b^2=ac}}$ $\Large \therefore x-2, x+2~ and ~x+4~in~GP$ $\Large \Rightarrow (x+2)^2=(x-2)(x+4)$ $\Large \Rightarrow 2x=-12$ $\Large \Rightarrow \color{forestgreen}{x=\boxed {-6}}$

Let the terms of the G.P. be $x-2,x+2,$ and $x+4$ , then the common ratio must be equal.

$\dfrac{x+2}{x-2}=\dfrac{x+4}{x+2}$

$(x+2)^2=(x-2)(x+4)$

$x^2+4x+4=x^2+4x-2x-8$

$12=-2x$

$\boxed{x=-6}$