Question

If $a,b,c$ are in $G.P.$ and $a^x=b^y=c^z$, prove that $\frac{1}{x}+\frac{1}{z}=\frac{2}{y}$

Answer 1

Since $a,b,c$ are in $G.P.$

$b^2=ac$ --- (i)

Now

$a^x=b^y=c^z=r$

$a=r^{\frac{1}{x}},b=r^{\frac{1}{y}},c=r^{\frac{1}{z}}$

From (i)

$r^{\frac{2}{y}}=r^{\frac{1}{x}+\frac{1}{z}}$

$\frac{1}{x}+\frac{1}{z}=\frac{2}{y}$