three numbers in GP are such that their product is 216 if the sum of product in pairs is 156 then the numbers are

• ans is 2,618
  a=2,r=3
• 10 years agoHelpfull: Yes(1) No(0)
• i) Let the 3 numbers in GP be a/r, a & ar
  
  ii) So their product is: (a/r)*(a)*(ar) = a^3 = 216; ==> a = 6
  
  iii) As such the 3 numbers are: 6/r, 6, 6r
  
  iv) As per the second constraint: (6/r)*(6) + 6*6r + (6r)*(6/r) = 156
  
  Taking LCM, simplifying and rearranging, 3r² - 10r + 3 = 0
  Factorizing, (3r - 1)(r - 3) = 0
  ==> Either r = 3 or r = 1/3
  
  As both are feasible solutions, substituting these in (iii), we get
  the 3 numbers as: {2, 6, 18} or {18, 6, 2}
  
  EDIT:
  In the final answer, the second triplet was {18, 6, 1}, which is now corrected as
  {18, 6, 2}
• 8 years agoHelpfull: Yes(0) No(0)