Question

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

Answer 1

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

So,

$b^2=ac$

$a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}$ (given)

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

then $a=k^x,b=k^y,c=k^z$

Now,

$b^2=ac$

$k^{2y}=k^x{k}^z$

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

$\Rightarrow 2y=x+z$

Therefore, $x,y,z$ are in A.P.