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The ninth term of the expansion $\left(3 x-\frac{1}{2 x}\right)^{8}$ is
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The correct answer is:
$\frac{1}{256 \cdot x^{8}}$
The general term of the expansion $(x+a)^{n}$ is $T_{r+1}={ }^{n} C_{r} x^{n-r} a^{r} .$
We have, $\left(3 x-\frac{1}{2 x}\right)^{8}$.
Here, $\quad r=8, x=3 x, a=\left(-\frac{1}{2 x}\right), n=8$
$\therefore$ Nineth term, $T_{9}={ }^{8} C_{8}(3 x)^{8-8}\left(\frac{-1}{2 x}\right)^{8}$
$$
=\frac{1}{256 \cdot x^{8}}
$$
We have, $\left(3 x-\frac{1}{2 x}\right)^{8}$.
Here, $\quad r=8, x=3 x, a=\left(-\frac{1}{2 x}\right), n=8$
$\therefore$ Nineth term, $T_{9}={ }^{8} C_{8}(3 x)^{8-8}\left(\frac{-1}{2 x}\right)^{8}$
$$
=\frac{1}{256 \cdot x^{8}}
$$
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