Explanation

(a)
Let $a^{1/x} = b^{1/y} = c^{1/\mathcal{z}} = k$
$\Rightarrow a = k^{x},\ b = k^{y},\ c = k^{\mathcal{z}}$
Now, $a,\ b,\ c$ are in GP.
$\Rightarrow \ b^{2} = ac\ \Rightarrow k^{2y} = k^{x} \bullet k^{\mathcal{z}} = k^{x + \mathcal{z}}$
$\Rightarrow \ 2y = x + \mathcal{z}$
$\therefore x,\ y,\ \mathcal{z}$ are in AP.