Solution :
Here, n = 20, which is an even number.
So, \({20\over 2} + 1\)th term i.e. 11th term is the middle term.
Hence, the middle term = \(T_{11}\)
\(T_{11}\) = \(T_{10 + 1}\) = \(^{20}C_{10}\) \(({2\over 3}x^2)^{20 – 10}\) \((-{3\over 2x})^{10}\)
= \(^{20}C_{10} x^{10}\)
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