Question

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

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

Answer 1

Answer: (C)

HP

$x^2+9y^2+25z^2=xyz(\frac{15}{x}+\frac{5}{y}+\frac{3}{z})$
$\Rightarrow x^2+9y^2+25z^2=3xy+15yz+5xz$
$\Rightarrow (x^2-6xy+9y^2)+(9y^2-30yz+25z^2)+(25z^2-10xz+x^2)=0$
$\Rightarrow {(x-3y)}^2+{(3y-5z)}^2+{(5z-x)}^2=0$
Therefore, $x-3y=0$; $3y-5z=0$; $5z-x=0$
$\Rightarrow x=3y=5z=k$
$\Rightarrow x=k$; $y=\frac{k}{3}$; $z=\frac{k}{5}$
now $\frac{2}{y}=\frac{6}{k}$ and $\frac{1}{x}+\frac{1}{z}=\frac{6}{k}$
$\frac{1}{x}+\frac{1}{z}=\frac{2}{y}$
Therefore, $x,y,z$ are in H.P

Ans: C