If $a, b, c$ are in GP and $a^{1 / x}=b^{1 / y}=c^{1 / z}$ then prove that $x, y, z$ are in AP.
It is given that
$a^{1 / x}=b^{1 / y}=c^{1 / z}$
Let $a^{1 / x}=b^{1 / y}=c^{1 / z}=k$
$\Rightarrow a^{1 / x}=k$
$\Rightarrow\left(a^{1 / x}\right)^{x}=k^{x} \ldots($ Taking power of $x$ on both sides. $)$
$\Rightarrow a^{1 / x \times x}=k^{x}$
$\Rightarrow a=k^{x}$
Similarly $b=k^{y}$
And $c=k^{z}$
It is given that a,b,c are in G.P.
$\Rightarrow \mathrm{b}^{2}=\mathrm{ac}$
Substituting values of a,b,c calculated above, we get:
$\Rightarrow\left(\mathrm{k}^{\mathrm{y}}\right)^{2}=\mathrm{k}^{\mathrm{x}} \mathrm{k}^{\mathrm{z}}$
$\Rightarrow \mathrm{k}^{2 \mathrm{y}}=\mathrm{k}^{\mathrm{x}+\mathrm{z}}$
Comparing the powers we get,
$2 y=x+z$
Which is the required condition for x,y,z to be in A.P.
Hence, proved that x,y,z, are in A.P.