Question

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

• A. $A.P.$
• B. $G.P.$
• C. $A.G.P.$
• D. $H.P.$

Answer 1

The correct option is D $H.P.$

Given, $x^2+9y^2+25z^2=15yz+5zx+3xy$
$(x{)}^2+(3y{)}^2+(5z{)}^2-(x\left)\right(3y)-(3y\left)\right(5z)-(x\left)\right(5z)=0$
$\frac{1}{2}\left[\begin{array}{c}2(x{)}^2+2(3y{)}^2+2(5z{)}^2-2(x\left)\right(3y)-2(3y\left)\right(5z)-2(x\left)\right(5z)\end{array}\right]=0$
$\frac{1}{2}\left[\begin{array}{c}(x-3y{)}^2+(3y-5z{)}^2+(x-5z{)}^2\end{array}\right]=0$
$\Rightarrow x-3y=0,3y-5z=0,x-5z=0$
$\frac{1}{x}=\frac{1}{3y}and\frac{5}{3y}=\frac{1}{z}and\frac{1}{5z}=\frac{1}{x}$
$\frac{1}{x}+\frac{1}{z}=\frac{1}{3y}+\frac{5}{3y}=\frac{2}{y}$
$\therefore \frac{1}{x},\frac{1}{y},\frac{1}{z}$
are in $A.P$. and $x,y,z$ are in $H.P.$