Answer: $\mathbf B$
Solution:
Let $a = 2$
and, $\mathrm{x = 4, y = 6, z = 8}$ be the terms in $\mathbf {AP}$ with common difference $=2$
Now,
$\mathrm{a^x = 4^2, a^y = 4^4, a^z = 4^6}$
$\mathrm {\frac{a^y}{a^x} = \frac{4^4}{4^2} = 16,\;\text{and} \;\frac{a^z}{a^y} = \frac{4^6}{4^4} = 16}$
So, the resultant terms are in $\mathbf GP$
$\therefore \mathbf B$ is the correct option.