If $x,y,z$ are in $A.P.$ and $a>1$, then $a^x, a^y, a^z$ are in
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.