Let (r + 1)th term be independent of in the given expression
Now Tr + 1 = 10Cr (3x)10 – r ( –\(\frac{1}{2x^3}\))r
= 10Cr (3x)10 – r ( –\(\frac{1}{2}\))r x30 − 6
This term is independent of x, if
30 – 6r = 0
⇒ r = 5
So, (5 + 1)th i.e 6th terms is independent of x