Let the sixth term in the binomial expansion of 2 log 2 10 - 3 x + 2 ( x - 2 ) log 2 3 5 m powers of 2 x - 2 log 2 3 , be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of x is _____ .

Step-by-step solution

Given, Binomial expression, 2 log 2 10 - 3 x + 2 ( x - 2 ) log 2 3 5 m 10 - 3 x + 3 ( x - 2 ) 5 m Now, T 6 = C 5 m 10 - 3 x m - 5 2 · 3 x - 2 = 21                   … 1 Also given, C 1 m , C 2 m , C 3 m are in A.P. So, 2 · C 2 m = C 1 m + C 3 m ⇒ 2 × m ! 2 ! m - 2 ! = m + m ! 3 ! m - 3 ! Solving for m , we get m = 2   ,   7 and m = 2 (rejected), so m = 7 Put in equation 1 21 · 10 - 3 x 3 x 9 = 21 ⇒ 10 - 3 x 3 x = 9 × 1