Find the middle term (terms) in the expansion of
(i) `((x)/(a) - (a)/(x))^(10)` (ii) `(3x - (x^(3))/(6))^(9)`

Explanation

(i) Given expansion is `((x)/(a) - (a)/(x))^(10)`
Here, the power of Binomial i.e., `n = 10` is even
Since, it has one middle term `((10)/(2) + 1)` th terms i.e., 6 th term
`:. T_(6) = T_(5 + 1) = .^(10)C_(5) ((x)/(a))^(10 - 5) ((-a)/(x))^(5)`
`= -.^(10)C_(5) ((x)/(a))^(5) ((a)/(x))^(5)`
`= - (10 xx 9 xx 8 xx 7 xx 6 xx 5!)/(5! xx 5 xx 4 xx 3 xx 2 xx 1) ((x)/(a))^(5) ((x)/(a))^(-5)`
`= - 9 xx 4 xx 7 = - 252`
(ii) given expansion is `(3x - (x^(3))/(6))^(9)`
Here, `n = 9`
Since, the Binomial expansion has two middle terms i.e., `((9 + 1)/(2)) th` and `((9 + 1)/(2) + 1)th` i.e., 5th term and 6th term.
`:. T_(5) = T_((4 + 1)) = .^(9)C_(4) (3x)^(9 - 4) (-(x^(3))/(6))^(4)`
`= (9 xx 8 xx 7 xx 6 xx 5!)/(4 xx 3 xx 2 xx 1 xx 5!) 3^(5) x^(5) x^(12) 6^(-4)`
`= (7 xx 6 xx 3 xx 3^(1))/(2^(4)) x^(17) = (189)/(8) x^(17)`
`:. T_(6) = T_(5) = .^(9)C_(5) (3x)^(9 - 5) (- (x^(3))/(6))^(5)`
`= - (9 xx 8 xx 7 xx 6 xx 5!)/(5! xx 4 xx 3 xx 2 xx 1) . 3^(4). x^(4). x^(15). 6^(-5)`
`= (-21 xx 6)/(3 xx 2^(5)) x^(19) = (-21)/(16) x^(19)`