Question

If $x,y,z$ are real and $4x^2+9y^2+16z^2-6xy-12yz-8zx=0$ , then $x,y,z$ are in

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

Answer 1

The correct option is C H.P.

Given,

$x,y,z$ are real

and

$4x^2+9y^2+16z^2-6xy-12yz-8zx=0$

$(2x{)}^2+(3y{)}^2+(4z{)}^2-(2x\left)\right(3y)-(3y\left)\right(4z)-(2x\left)\right(4z)=0$

[$\frac{1}{2}\times$ $(2x-3y{)}^2+(3y-4z{)}^2+(4z-2x{)}^2]=0$

$(2x-3y{)}^2+(3y-4z{)}^2+(4z-2x{)}^2=0$

sum of square can not be $0$ so this means these all are $0$

$2x=3y$ and $3y=4z$

$2x=3y=4z=k$

so , $x=\frac{k}{2}$ , $y=\frac{k}{3}$ and $z=\frac{k}{4}$

So, $x,y,z$ are in $H.P$