If both the roots of the quadratic equation $(b-c)x^2+(c-a)x+(a-b)=0$ are equal. Then $a, b, c$ are in

$\begin{array}{1 1} AP \\ GP \\ HP \\ No\;such\;relationship\;exists \end{array}$

Explanation : since both roots are equal,In a quadratic equation of type $lx^2+mx+n=0$, roots are equal if $m^2-4ln=0$
$(c-a)^2-4(b-c)(a-b)=0$
$c^2-2ac+a^2+4b^2+4ac-4ab-4bc=0$
$c^2+a^2+4b^2+2ac-4ab-4bc=0$
$(c+a-2b)^2=0$
$c+a-2b=0$
$c+a=2b$
a,b,c are in AP.