Solve the following


If $r^{\text {th }}$ term is the middle term in the expansion of $\left(x^{2}-\frac{1}{2 x}\right)^{20}$, then $(r+3)^{t h}$ term is

(a) ${ }^{20} C_{14}\left(\frac{x}{2^{14}}\right)$

(b) ${ }^{20} C_{12} x^{2} 2^{-12}$

(c) ${ }^{20} C_{7} x, 2^{-13}$

(d) none of these


(c) ${ }^{20} C_{7} x, 2^{-13}$

Here $n$ is even

So, The middle term in the given expansion is $\left(\frac{20}{2}+1\right)$ th $=11$ th term

Therefore, $(r+3)$ th term is the 14 th term.

$T_{14}={ }^{20} C_{13}\left(x^{2}\right)^{20-13}\left(\frac{-1}{2 x}\right)^{13}$

$=(-1)^{13}{ }^{20} C_{13} \frac{x^{14-13}}{2^{13}}$

$=-{ }^{20} C_{7} x 2^{-13}$

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