Respuesta :

[tex]\text{Let,}\\\\~~~~~~~a^x =b^y = c^z = k\\\\\text{So,}\\ \\a = k^{\dfrac 1x}\\\\b=k^{\tfrac 1y}\\\\\ c = k^{\tfrac1z}\\\\\text{Now, given that,}\\\\~~~~~~~~~b^2 = ac\\\\\implies \left( k^\tfrac 1y \right)^2 = k^{\tfrac 1x} \cdot k^{\tfrac 1z}\\\\\implies k^{\tfrac 2y} = k^{\tfrac 1x + \tfrac 1z} ~~~~~~~~~~;[a^m \cdot a^n = a^{m+n}]\\\\\implies \dfrac 2y = \dfrac 1x + \dfrac 1z\\\\\implies \dfrac 1x + \dfrac 1z = \dfrac 2y \\\\\text{Proved.}[/tex]