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Q.
$6^{th}$ term in expansion of $\left(2 x^{2}-\frac{1}{3 x^{2}}\right)^{10}$ is:
Jharkhand CECEJharkhand CECE 2003
Solution:
The general term in the expansion of $(x+a)^{n}$ is
$T_{r+1}={ }^{n} C_{r} x^{n-r} d^{r}$
Given expansion is $\left(2 x^{2}-\frac{1}{3 x^{2}}\right)^{10} $
$\therefore T_{r+1}={ }^{10} C_{r}\left(2 x^{2}\right)^{10-r}\left(-\frac{1}{3 x^{2}}\right)^{r} $
$\Rightarrow T_{6}={ }^{10} C_{5}\left(2 x^{2}\right)^{5}\left(-\frac{1}{3 x^{2}}\right)^{5}$
$=-\frac{10 ! \times 32}{5 ! 5 !} \times \frac{1}{243}=-\frac{896}{27}$