If the ${\mathrm{k}}^{\mathrm{th}}$ term in the expansion of ${\left(\frac{3}{2}{\mathrm{x}}^2-\frac{1}{3\mathrm{x}}\right)}^6$is independent of x, then the numerically greatest term in the expansion of ${\left(\frac{3}{2}{\mathrm{x}}^2-\frac{1}{3\mathrm{x}}\right)}^{\mathrm{k}}$ when $\mathrm{x}=\frac{1}{2},$ is

$$\begin{gathered} {\left(\frac{3}{2}{\mathrm{x}}^2-\frac{1}{3\mathrm{x}}\right)}^{\mathrm{k}}={\left(\frac{3}{2}{\mathrm{x}}^2-\frac{1}{3\mathrm{x}}\right)}^5 \\ \mathrm{For} \mathrm{N}.\mathrm{G}.\mathrm{T} \frac{(\mathrm{n}+1) \left(\mathrm{b}\right)}{\left|\mathrm{a}\right|+\left|\mathrm{b}\right|}=\frac{\left(6\right) \left({\displaystyle \frac{1}{3\mathrm{x}}}\right)}{{\displaystyle \frac{3}{2}}{\mathrm{x}}^2+{\displaystyle \frac{1}{3\mathrm{x}}}}=\frac{6\left({\displaystyle \frac{1}{2}}\right)}{{\displaystyle \frac{2}{3}}+{\displaystyle \frac{1}{2}}} \\ =\frac{3}{\left(7\right)}\left(6\right)=\frac{18}{7}=2.5 \\ {\mathrm{T}}_3 \mathrm{is} \mathrm{N}.\mathrm{G}.\mathrm{T} \\ {\mathrm{T}}_3={}^5{\mathrm{C}}_2 {\left(\frac{3}{2}{\mathrm{x}}^2\right)}^3 {\left(\frac{-1}{3\mathrm{x}}\right)}^2 \\ =\left(10\right) {\left(\frac{3}{2}\times \frac{4}{9}\right)}^3 {\left(\frac{1}{2}\right)}^2=10\times \frac{8}{27}\times \frac{1}{4} \\ =\frac{20}{27} \end{gathered}$$