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

  1. $A.P.$
  2. $G.P$
  3. $H.P$
  4. none of these
in Numerical Ability
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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.


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