Answers:

(i) 

Given,

(2/3x – 3/2x)²⁰  [n = 20]

Middle term –

(n/2 + 1) = (20/2 + 1) = (10 + 1) = 11.

The term is 11^th term

T₁₁ = T₁₀₊₁

T₁₁ = ²⁰C₁₀ (2/3x)^20-10 (3/2x)¹⁰

T₁₁ = ²⁰C₁₀ 2¹⁰/3¹⁰ × 3¹⁰/2¹⁰ x^10-10

T₁₁ = ²⁰C₁₀

Middle term is ²⁰C₁₀.

(ii)

Given,

(a/x + bx)¹²  [n = 12]

Middle term –

(n/2 + 1) = (12/2 + 1) = (6 + 1) = 7.

The term is 7^th term

T₇ = T₆₊₁

= 924 a⁶b⁶

Middle term is 924 a⁶b⁶.