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?
 A: The mean of two distinct reals lies always in between of them on the number line. We consider three cases:
Case 1: $y <z $
Then $y <x<z $ follows but since $y$ is the geometric mean of $x$ and $z$ it must lie between them. Contradiction!
Case 2: $y>z$
Then $y>x>z $ follows but since $y$ is the geometric mean of $x$ and $z$ it must lie between them. Contradiction!
Case 3: $y=z $
Then $x=y=z$ follows which satisfies all the given conditions (as one can easily check).
Hence, $x $, $y $ and $z $ must satisfy $x=y=z$, otherwise there will be a contradiction. If three equal numbers are in arithmetic, geometric or harmonic progression, that's your choice. 
In my opinion, they are in arithmetic progression since $y=x+0$ and $z=x+2 \cdot 0$. But for me, they are also in harmonic and geometric progression where you can argue analogously.
EDIT: Unfortunately, this solution only holds for $x, y, z\geq0$. See the other answers for a complete solution.
A: HINT:
We have $$y+z=2x=2\cdot\dfrac{y^2}z\implies0=z^2+yz-2y^2=(z-y)(z+2y)$$
A: 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.
