If the equation (m-n)x2+(n-l)x+l-m=0 has equal roots, then l, m and n satisfy

Roots are equal so b2-4ac=0⇒(n-l)2-4(m-n)(l-m)=0⇒n2+l2-2nl-4(ml-nl-m2+mn)=0⇒n2+l2-2nl-4ml+4nl+4m2-4mn=0⇒l2+n2+(2m)2+2nl-4mn-4ml=0⇒(l+n-2m)2=0⇒l+n=2m ⇒l, m, n are in A.P.