If x, y, z are real and 4x2 + 9y2 + 16z2 - 6xy - 12yz - 8zx = 0, then x, y, z are in

Question

If x, y, z are real and 4x2 + 9y2 + 16z2 - 6xy - 12yz - 8zx = 0, then x, y, z are in (A) A.P. (B) G.P. (C) H.P. (D) none of these

Tardigrade Admin · Accepted Answer

Multiplying the given expression by 2 and rewriting it, we have (2x-3y)2 +(3y -4z)2 +(4z -2x)2 =0 ⇒ 2x=3y=4z ⇒ (1/x),(1/y),(1/z) are in A.P ⇒ x, y, x are in H.P.

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

A.P.

G.P.

H.P.

none of these

Multiplying the given expression by 2 and rewriting it, we have
$(2 x - 3 y)^{2} + (3 y - 4 z)^{2} + (4 z - 2 x)^{2} = 0$
$\Rightarrow 2 x = 3 y = 4 z$
$\Rightarrow \frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P
$\Rightarrow x, y, x$ are in H.P.

Q. If x, y, z are real and 4x2+9y2+16z2−6xy−12yz−8zx=0, then x, y, z are in