If $1,{\log }_{y}x,{\log }_{z}y,-15{\log }_{x}z$  are in A.P, then

Let $d$  be the common differences. Then. ${\log }_{y}x=1+d\Rightarrow x=y^{1+d}$ 

            ${\log }_{z}y=1+2d\Rightarrow y=z^{1+2d}$

and     $-15{\log }_{x}z=1+3d\Rightarrow z=x^{-\left(1+3d\right)/15}$

$\therefore x=y^{1+d}=z^{\left(1+2d\right)\left(1+d\right)}$     $=x^{-\left(1+d\right)\left(1+2d\right)\left(1+3d\right)/15}$

$\Rightarrow \left(1+d\right)\left(1+2d\right)\left(1+3d\right)=-15$

$\Rightarrow 6d^3+11d^2+6d+16=0$

$\Rightarrow \left(d+2\right)\left(6d^2-d+8\right)=0\Rightarrow d=-2$

$\therefore x=y^{1+d}=y^{-1},y=z^{1-4}=z^{-3}$

and            $x={\left(z^{-3}\right)}^{-1}=z^3$