If $a, b, c$ are in $G.P.$ and $a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} = k,$ prove that $x, y, z$ are in $A.P.$

(N/A) Let $a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} = k.$
Then $a = k^{x}, b = k^{y},$ and $c = k^{z}$ $(1).$
Since $a, b, c$ are in $G.P.,$ we have $b^{2} = ac$ $(2).$
Substituting $(1)$ into $(2),$ we get $(k^{y})^{2} = k^{x} \cdot k^{z}.$
This simplifies to $k^{2y} = k^{x+z}.$
Equating the exponents,we get $2y = x + z.$
Since $2y = x + z,$ it follows that $x, y, z$ are in $A.P.$