Question

If $a,b,c$ are in $GP$ and $a^{1/x}=c^{1/z}$, then $x,y,z$ are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. $None$

Answer 1

The correct option is A $A.P.$

Given that,

$a,b,c$ in G.P.

then,

$b^2=ac$ ……….$\left(1\right)$

and

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

let $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}=k$

$a^{\frac{1}{x}}=k$

$b^{\frac{1}{y}}=k$

$b=k^y$

$c^{\frac{1}{z}}=k$

$c=k^2$

By equation $\left(1\right)$

$b^2=ac$

$(k^y{)}^2=k^x+k^z$

$k^{2y}=k^{x+z}$

On comparing and we get

$2y=x+z$

Hence, x, y, z are in an A.P.