If a, b, c are in H.P., then the roots of the equation a(b−c)x2+b(c−a)x+c(a−b)=0

We know, that a (b -c) +b (c -a) + c (a-b) =0. So, x =1 is a root of the given equation. Let α be the other root. Then α×1=c(a−b)a(b−c)⇒α=1b−1a1c−1b⇒ α=1 ∵a,b,care in H.P.∴1a,1b,1care in A.P.Thus, the given equation has equal roots each equal to 1.