If $x$ be the $A.M$ between $y$ and$z$... If $x$ be the AM between $y$ and $z$, $y$ be the GM between $z$ and $x$, then $x$, $y$, $z$ are in :
$1$). A.P
$2$). G.P
$3$). H.P
$4$). None.
My Attempt:
$x$ is the AM between $y$ and $z$
$$x=\dfrac {y+z}{2}$$
$$2x=y+z$$
$y$ is the G.M between $z$ and $x$.
$$y=\sqrt {zx}$$
$$y^2=xz$$
What should I do next?

In my opinion... They are in A.P, G.P and H.P too.. let's have a look:
According to your question, you can have $2x=y+z$ and $xz=y^2$.
These two equations give you $\frac{2x}{xz}=\frac{y+z}{y^2} \implies \frac{2}{z}=\frac{y+z}{y^2} \implies z^2+yx-2y^2=0 \implies (z-y)(z+2y)=0 $.
Now we have $z=y$ or $z=-2y$. The first one implies that $x,y,z$ are in A.P. or H.P and the second one implies that they are in G.P.
