If the seventh, terms from the beginning and the end in the expansion of `(root(3)(2)+1/(root(3)(3)))^(n)` are equal, then n is equal to
(i) 10
(ii) 11
(iii) 12
(iv) 13

Explanation

Given expansions is `(root(3)(2) + (1)/(root(3)(3)))^(n)`
`:. T_(7) = T_(6 + 1) = .^(n)C_(6) (root(3)(2))^(n - 6) ((1)/(root(3)(3)))^(6)`
Since, `T_(7)` from end is same as the `T_(7)` from beginning of `((1)/(root(3)(3))) + ((1)/(root(3)(3)) + root(3)(2))^(n)`
Then, `T_(7) = .^(n)C_(6) ((1)/(root(3)(3)))^(n - 6) (root(3)(2))^(6)`
Given, that, `.^(n)C_(6)(2)^((n -6)/(3)) (3)^(-6//3) = .^(n)C_(6) (3)^(-((n - 6))/(3)) 2^(6//3)`
`rArr (2)^((n - 12)/(3)) = ((1)/(3^(1//3)))^(n - 12)`
Which is true, when `(n - 12)/(3) = 0`
`rArr n - 12 0 rArr n = 12`