Question

If a, b, c are in G.P. and $a^{\frac{1}{x}=b^{\frac{1}{y}}=c^{\frac{1}{z}}}$

• A. AP
• B. GP
• C. HP
• D. none of these

Answer 1

The correct option is A

AP

(a) AP

a, b and c are in G.P.

$\therefore b^2=ac$

Taking log on both the sides :

$2logb=loga+logc$ .......... (i)

Now, $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}$

Taking log on both the sides :

$\frac{loga}{x}=\frac{logb}{y}=\frac{logc}{z}$ ....... (ii)

Now, comparing (i) and (ii) :

$\frac{loga}{x}=\frac{loga+logc}{2y}=\frac{logc}{z}$

$\Rightarrow \frac{loga}{x}=\frac{loga+logc}{2y}$ and $\frac{loga}{x}=\frac{logc}{z}$

$\Rightarrow loga(2y-x)=xlogcand\frac{loga}{logc}=\frac{x}{z}$

$\Rightarrow \frac{loga}{logc}=\frac{x}{(2y-x)}$ and $\frac{loga}{logc}=\frac{x}{z}$

$\Rightarrow \frac{x}{(2y-x)}=\frac{x}{z}$

$\Rightarrow 2y=x+z$
Thus, x, y and z are in A.P.