Find the 4th term from the beginning and 4th term from the end in the expansion of

Let Tr+1 be the 4th term from the end.

Then, Tr+1 is (10 − 4 + 1)th, i.e., 7th, term from the beginning.

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

$={ }^{9} C_{6}\left(x^{9-6}\right)\left(\frac{2}{x}\right)^{6}$

$=\frac{9 \times 8 \times 7}{3 \times 2}\left(x^{3}\right)\left(\frac{64}{x^{6}}\right)$

$=\frac{5376}{x^{3}}$

4th term from the beginning $=T_{4}=T_{3+1}$

$\therefore T_{4}={ }^{9} C_{3}\left(x^{9-3}\right)\left(\frac{2}{x}\right)^{3}$

$=\frac{9 \times 8 \times 7}{3 \times 2}\left(x^{6}\right)\left(\frac{8}{x^{3}}\right)$

$=672 x^{3}$