Question

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

• A. x, y, z are in G.P.
• B. x, y, 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)

$4^x=16^y=64^z=K$
$\Rightarrow 4=K^{\frac{1}{x}}:16=K^{\frac{1}{y}},64=K^{\frac{1}{z}}$
$\Rightarrow 4\times 64=K^{\frac{1}{x}}:K^{\frac{1}{z}}=K^{\frac{2}{y}}=K^{\frac{1}{x}+\frac{1}{z}}$
$\Rightarrow 256=K^{\frac{1}{x}}\cdot K^{\frac{1}{z}}=K^{\frac{2}{y}}=K^{\frac{1}{x}+\frac{1}{z}}$
$\Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z}$
$\Rightarrow \frac{1}{x}\cdot \frac{1}{y}\cdot \frac{1}{z}$ are in $A.P$.
$\left(\because 256=16^2=K^{\frac{2}{y}}\right)$