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 these

• Suppose the roots are r and s. Then (x-r)(x-s) = 0, so
  x^2 - (r+s) x + (rs) = 0
  and hence r+s = -b/a and rs = c/a.
  
  We are saying that
  r + s = (1/r^2 + 1/s^2)
  (rs)^2 (r+s) = (rs)^2 (1/r^2 + 1/s^2) = s^2 + r^2
  (rs)^2 (r+s) = (r+s)^2 - 2rs
  (c/a)^2 (-b/a) = (-b/a)^2 - 2(c/a)
  Multiplying through by a^3 we get
  -bc^2 = ab^2 - 2a^2c
  Divide through by abc to get
  -c/a = b/c - 2a/b
  or a/b - c/a = b/c - a/b
  which means c/a, a/b and b/c are in AP.
  
  Hence their reciprocals, a/c, b/a and c/b are in harmonic progression (HP).
• 12 years agoHelpfull: Yes(12) No(2)
• in HP becoz c/a , a/b , b/c are in AP after solving equations
  so answer is
  (c)HP
• 12 years agoHelpfull: Yes(3) No(3)