Answers

2016-03-30T10:47:36+05:30
Let the numbers be  a, ar, ar²
 Now, sum of the numbers :   a+ar+ar² =  57  ----------- (i)
    "     product of numbers :    a*ar*ar² = 343  -----------(ii)
                                           a³r³ = 343
                                           (ar)³ = 343
     on comparing, we get 
                                         ar = 7 ⇒ a = 7/r
         putting the value of ar in eq. (i)
 7/r + 7 + 7r = 57
  7r² - 50r - 7 = 0
  7r² -49r - r - 7 = 0
 7r (r-7) +1 (r-7) = 0
∴  r = 7    &      a = 1
Hence, numbers are   1, 7, 49
0