AP and GP

We are given that $a$ , $b$ and $c$ are in an arithmetic progression with common difference $6$

$a=x \quad b=x+6 \quad c=x+12$

$b$ , $c$ and $d$ are in a geometric progression, and its common ratio is

$r=\dfrac{c}{b}=\dfrac{x+12}{x+6}$

Which implies that

$d=rc=\dfrac{(x+12)^2}{x+6}$

But we are also given that $a=d$ , so

$\begin{aligned} a=d&\iff x=\dfrac{(x+12)^2}{x+6}\\ &\iff x(x+6)=(x+12)^2\\ &\iff x^2+6x=x^2+24x+144\\ &\iff 18x=-144 \iff x=-8 \end{aligned}$

To conclude

$(a,b,c,d)=(-8,-2,4,-8)$

And thus

$a+b+c+d=-8-2+4-8=\boxed{-14}$