Question

The middle term in the expansion of ${\left(\frac{2x^2}{3}+\frac{3}{2x^2}\right)}^{10}$ is
(a) 251
(b) 252
(c) 250
(d) none of these

Answer 1

(b) 252

$$\begin{gathered} \mathrm{Here},n,\mathrm{i}.\mathrm{e}.,10,\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{number}. \\ \therefore \mathrm{Middle}\mathrm{term}=\left(\frac{10}{2}+1\right)\mathrm{th}\mathrm{term}=6\mathrm{th}\mathrm{term} \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ {T}_6=T_{5+1} \\ ={}^{10}C_5{\left(\frac{2x^2}{3}\right)}^{10-5}{\left(\frac{3}{2x^2}\right)}^5 \\ =\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\times \frac{2^5}{3^5}\times \frac{3^5}{2^5} \\ =252 \end{gathered}$$