Question

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

• A. $x,y,z$ are in H.P.
• B. $x=2z$
• C. $x,y,z$ are in A.G.P.
• D. $2x=3y$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $x,y,z$ are in H.P.
B $x=2z$
D $2x=3y$

$4x^2+9y^2+16z^2-6xy-12yz-8zx=0...\left(1\right)$
$\Rightarrow 4x^2-2x(3y+4z)+9y^2+16z^2-12yz=0$

Since, $x\in R\Rightarrow D\ge 0$
$\Rightarrow 4(3y+4z{)}^2-4\cdot 4\cdot (9y^2+16z^2-12yz)\ge 0$
$\Rightarrow 27y^2+48z^2-72yz\le 0$
$\Rightarrow 9y^2+16z^2-24yz\le 0$
$\Rightarrow (3y-4z{)}^2\le 0$
$\Rightarrow 3y=4z$

Substituting in given equation $\left(1\right)$
$\Rightarrow 4x^2+18y^2-6xy-9y^2-6xy=0$
$\Rightarrow 4x^2+9y^2-12xy=0$
$\Rightarrow (2x-3y{)}^2=0$
$\Rightarrow 2x=3y$
$\therefore 2x=3y=4z=\lambda$
$\Rightarrow x=\frac{\lambda }{2},y=\frac{\lambda }{3},z=\frac{\lambda }{4}$
Hence, $x,y,z$ are in H.P.