Question

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

Answer 1

Let,

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

Then

$a=k^x,b=k^y,c=k^z$...(1)

Since, a,b,c are in GP

$\therefore b^2=ac$...(2)

Using (1) in (2) we get,

$k^{2y}=k^{x+z}$ which gives $2y=x+z$

Hence, x, y and z are in AP.