If $x$ be the $A.M$ between $y$ and$z$...

Question

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?

Answer 1

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.

Answer 2

HINT:

We have $$y+z=2x=2\cdot\dfrac{y^2}z\implies0=z^2+yz-2y^2=(z-y)(z+2y)$$

Answer 3

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.