$\text{ The numerically greatest term of }(2+3x{)}^8\text{ when }x=\frac{1}{2}$

${(2+3x)}^8={\left(2\right)}^8{\left(1+\frac{3x}{2}\right)}^8$

$\text{ Here }n=8;X=\frac{3x}{2}$

$\text{ When }x=\frac{1}{2},X=\frac{3}{4}$

now

$\begin{array}{l}\frac{(n+1)\left|X\right|}{\left|X\right|+1}=\frac{(8+1)\times \frac{3}{4}}{\frac{3}{4}+1}\\ =\frac{9\times \frac{3}{4}}{\frac{7}{4}}\\ =\frac{27}{7}\\ =3.8\\ =3+0.8\end{array}$

                       = p+F

$\therefore {\text{ Numerically greatest term is=T}}_{p+1}\text{= }{T}_{3+1}=T_4\left(4^{\text{th }}\text{ term }\right)$