Question

If $l(m-n)x^2+m(n-l)xy+n(l-m)y^2$ is a perfect square, then the quantities $l,m,n$ are in

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

Answer 1

The correct option is B H.P.

Since, $l(m-n)x^2+m(n-l)xy+n(l-m)y^2$ is a perfect square.
Therefore, $l(m-n)x^2+m(n-l)xy+m(l-m)y^2=0$ are equal.
$\therefore D={\left[m\right(n-l\left)\right]}^2-4ln(m-n)(l-m)=0$
$\Rightarrow m^2{(n-l)}^2=-4ln[m^2-m(l+n)+ln]$
$\Rightarrow m^2[{(n-l)}^2+4ln]-4lmn(l+n)+4l^2{n}^2=0$
$\Rightarrow m^2{(l+n)}^2-4lmn(l+n)+4l^2{n}^2=0$
$\Rightarrow {[m(l+n)-2ln]}^2=0$
$\Rightarrow m(l+n)=2ln$
$\Rightarrow m=\frac{2ln}{l+n}$
Therefore, $l,m,n$ are in H.P
Ans: B