If rth term is the middle term in the expansion of (x2−12x)20, then (r+3)th term is
−20C7x.2−13
−20C7x.2−13
Here n is even
So, The middle term in the given expansion is (202+1)th =11th term
Therefore, (r+3)th term is the 14th term
T14=20C13(x2)20−13(−12x)13
=(−1)1320C13x14−13213
=−20C7x2−13