If ( 12 - x ) , 12 , ( 12 + x ) in order are three consecutive terms of a G.P. then sum of all the solutions in [0, 314] is k . The value of k is:

Step-by-step solution

Since they are in G.P., we have: ^2 12 = ( 12 - x ) ( 12 + x ) ^2 12 = ^2 12 - ^2 x 1 - ^2 12 ^2 x Let t = ^2 12 and y = ^2 x . t = t - y 1 - ty t - t^2y = t - y y(1 - t^2) = 0 . So ^2 x (1 - ^4 12 ) = 0 . Since 12 1 , we must have ^2 x = 0 x = 0 . The solutions are x = n , where n Z . In the interval [0, 314] , since 3.14159 , 100 314.159 , so the solutions are x = 0, , 2 , , 99 . The sum is 0 + + 2 + + 99 = 99 100 2 = 4950 . Hence, k = 4950 .