Find the 5th term from the end in the expansion

Given:

$\left(3 x-\frac{1}{x^{2}}\right)^{10}$

Clearly, the expression has 6 terms.

The 5th term from the end is the (11 − 5 + 1)th, i.e., 7th, term from the beginning.

Thus, we have:

$T_{7}=T_{6+1}$

$={ }^{10} C_{6}(3 x)^{10-6}\left(\frac{-1}{x^{2}}\right)^{6}$

$={ }^{10} C_{6}\left(3^{4}\right)\left(x^{4}\right)\left(\frac{1}{x^{12}}\right)$

$=\frac{10 \times 9 \times 8 \times 7 \times 81}{4 \times 3 \times 2 \times 1 \times x^{8}}=\frac{17010}{x^{8}}$