Find the 5th term from the end in the expansion of (x³/2 − 2/x²)⁹.

We know that the p^th term from the end = (n − p + 2)^th term from the beginning.

∴ In the expansion of (x³/2 − 2/x²)⁹, we have the 5th term from the end

= (9 − 5 + 2)^th term = 6^th term from the beginning = T₆ = T₅₊₁

= (−1)⁵ × ⁹C₅ × (x³/2)⁹⁻⁵ × (2/x²)⁵

= −⁹C₅ × (x³/2)⁴ × (2/x²)⁵

= −(9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) × (x¹²/16) × (32/x¹⁰)

= −252x²

Hence, the 5th term from the end is −252x².