Question

If $x^2+9y^2+25z^2=xyz(\frac{15}{x}+\frac{5}{y}+\frac{3}{z})$ where x, y, z $\in R$, then x, y, z satisfy.

• A. $2y=x+z$
• B. $y^2=xz$
• C. $\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$
• D. $y^2=x^2{z}^2$

Answer 1

Answer: (C)

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

$x^2+9y^2+25z^2=xyz(\frac{15}{x}+\frac{5}{y}+\frac{3}{z})$

$x^2+9y^2+25z^2=15xy+5xz+3xy$

$⟹x^2+9y^2+25z^2-15xy-5xz-3xy=0$

$\frac{1}{2}\left(\right(x-3y{)}^2+(3y-5z{)}^2+(5z-x{)}^2)=0$

Three square terms are equal to $0$. Its possible when

$x=3y,3y=5z,x=5z$

OR

$x=3y=5z=k$

$x=k,y=k/3,z=k/5$

Stisfying the equation

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