Question

If ${{4}^{x}}={{16}^{y}}={{64}^{z}},$ then

• A. $x,\text{ }y,\text{ }z$ are in $G.P.$
• B. $x,\text{ }y,\text{ }z$ are in $A.P.$
• C. $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in $G.P.$
• D. $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in $A.P.$

Answer 1

Answer: (D)

$\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in $A.P.$

Answer 2

Given, ${{4}^{x}}={{16}^{y}}={{64}^{z}}$
$\Rightarrow$ ${{2}^{2x}}={{2}^{4y}}={{2}^{6z}}$
$\Rightarrow$ $2x=4y=6z$
$\Rightarrow$ $x=2y=3z$
$\Rightarrow$ $\frac{x}{6}=\frac{y}{3}=\frac{z}{2}=k$
$\Rightarrow$ $x=6k,\,y=3k,\,z=2k$
Let $\frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in AP.
$\Rightarrow$ $\frac{1}{6k},\frac{1}{3k},\frac{1}{2k}$ are in AP.
$\Rightarrow$ $2.\frac{1}{3k}=\frac{1}{6k}+\frac{1}{2k}=\frac{4}{6k}$
$\Rightarrow$ $\frac{2}{3k}=\frac{2}{3k}$