Question

If $a^x=b^y=c^z$ and $b^2=ac$, then $\frac{1}{x}$ + $\frac{1}{z}$

• A. $\frac{2}{y}$
• B. $\frac{1}{y}$
• C. $\frac{1}{2}y$
• D. $2y$

Answer 1

The correct option is A $\frac{2}{y}$

$a^x=b^y$

$=>a=b^{\frac{y}{x}}$ (i)

Also,

$c^z=b^y$

$=>c=b^{\frac{y}{z}}$ (ii)

Now,

$b^2=ac$

$=>b^2=b^{\frac{y}{x}}{b}^{\frac{y}{z}}$ using (i) (ii)

$=>b^2=b(\frac{y}{x}+\frac{y}{z})$

$=>b^2=b^y(\frac{1}{x}+\frac{1}{z})$

Thus,

$2=y(\frac{1}{x}+\frac{1}{z})$

$=>\frac{1}{x}+\frac{1}{z}=\frac{2}{y}$