Question

If a, b, c are in G.P. and $a^x=b^y=c^z$ , then prove that $\frac{1}{x}+\frac{1}{z}=\frac{2}{y}.$

Answer 1

$a^x=b^y=c^z$

$a,b,c\to GP$

$b^2=ac$

$xloga=ylogb=zlogc$

$xloga=\frac{y}{2}logac$

$xloga=\frac{y}{2}(loga+logc)$

$(2x-y)loga=ylogc$

$\frac{loga}{logc}=\frac{y}{2x-y}$ __ (1)

$\frac{y}{2}logac=zlogc$

$\frac{y}{2}(loga+logc)=zlogc$

$yloga=(2z-y)logc$

$\frac{loga}{logc}=\frac{2z-y}{y}$ __ (2)

(1) = (2)

$\frac{y}{2x-y}=\frac{2z-y}{y}$

$y^2=(2x-y)(2z-y)$

$4\times z-y(2x+2z)=0$

$2\times 2-y(x+2)=0$

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