If $a^{1/x} = b^{1/y} = c^{1/z}$ and $a, b, c$ are in $G.P.$,then $x, y, z$ will be in

Let $0 < z < y < x$ be three real numbers such that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression and $x, \sqrt{2}y, z$ are in a geometric progression. If $xy + yz + zx = \frac{3}{\sqrt{2}} xyz$,then $3(x + y + z)^2$ is equal to $............$.

The arithmetic mean and the geometric mean of two distinct $2$-digit numbers $x$ and $y$ are two integers,one of which can be obtained by reversing the digits of the other (in base $10$ representation). Then,$x+y$ equals

If $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ such that $a < b < c$ and $a+b+c = \frac{3}{4}$,then the value of $a$ is

If $\alpha, \beta, \gamma$ are the geometric means between $ca, ab$; $ab, bc$; and $bc, ca$ respectively,where $a, b, c$ are in $A.P.$,then $\alpha^2, \beta^2, \gamma^2$ are in

If $a, b, c$ are in $A.P.;$ $b, c, d$ are in $G.P.$ and $\frac{1}{c}, \frac{1}{d}, \frac{1}{e}$ are in $A.P.,$ prove that $a, c, e$ are in $G.P.$