If the sum of the roots of the equation ax²+bx+c=0 is equal to the sum of the squares of their reciprocals,
then, a/c, b/a, c/b are in

(a) AP
(b) GP
(c) HP
(d) None of the these

• a/c, b/a, c/b are in HP... option C
• 11 years agoHelpfull: Yes(10) No(2)
• Let roots be P and q
  then p+q=(1/p^2)+(1/q^2)=(p^2+q^2)/(P^2*q^2)
  p+q = ((p+q)^2 - 2*p*q)/(p*q)^2-----------------(1)
  we know that sum of roots p+q = -b/a
  and product of roots p*q = c/a
  putting these values in eqn (1), we get-
  -b/a = ((-b/a)^2 - 2*c/a)/(c/a)^2
  (-b/a)*(c^2/a^2)= (b^2/a^2)- 2*c/a
  multiplying both sides with a^3,we get-
  a*b^2 - 2*c*a^2 = -b*c^2
  dividing both sides again by abc, we get--
  2*a/b = b/c + c/a
  b/(2*a) = a*c/(ab + c^2)
  b/a = (2*(a/c)*(c/b))/(a/c + c/b)-----------------(2)
  we know that three no. k,l,m are in h.p. if l = (2*k*m)/(k+m)
  So from eqn (2), we can say that a/c,b/a,c/b are in h.p.
  so ANSWER = option (c)
  
• 11 years agoHelpfull: Yes(7) No(2)
• i get answer b^2=a^2.
  but i dont knw they r in a.p r g.p r h.p?
  anyone can explain?
  
• 11 years agoHelpfull: Yes(1) No(5)
• ap option a
• 11 years agoHelpfull: Yes(0) No(10)
• Answer is HP.
• 7 years agoHelpfull: Yes(0) No(0)