Given that 4th term in the expansion of  ${(2+\frac{3x}{8})}^{10}$ has maximum numerical value
then ‘x’ lies in the interval
 

     ${(2+\frac{3x}{8})}^{10}=2^{10}{(1+\frac{3x}{16})}^{10}$     in the given expansion 4th term is the numerically greatest term
We have  $3<\frac{11\times \frac{3}{16}\left|x\right|}{1+\frac{3}{16}\left|x\right|}<4\Rightarrow 3<\frac{33\left|x\right|}{16+3\left|x\right|}<4$
 $$\begin{gathered} 3<\frac{33\left|x\right|}{16+3\left|x\right|}\Rightarrow \left|x\right|>2\Rightarrow x<-2orx>2 \\ and\frac{33\left|x\right|}{16+3\left|x\right|}<4\Rightarrow \left|x\right|<\frac{64}{21}\Rightarrow -\frac{64}{21}<x<\frac{64}{21} \end{gathered}$$
$\therefore x\in (-\frac{64}{21},-2)\cup (2,\frac{64}{21})$