If the fourth term in the expansion of  ${\left(\mathrm{ax}+\frac{1}{\mathrm{x}}\right)}^{\mathrm{n}}$ is $\frac{5}{2},$ then
 

$\begin{array}{rcl}{\mathrm{T}}_4& =& {}^{\mathrm{n}}{\mathrm{C}}_3 {(\mathrm{a} \mathrm{x})}^{\mathrm{n}-3} {\left(\frac{1}{\mathrm{x}}\right)}^3 \\ & =& {}^{\mathrm{n}}{\mathrm{C}}_3 {\mathrm{a}}^{\mathrm{n}-3} {\mathrm{x}}^{\mathrm{n}-6}\end{array}$

Since  ${\mathrm{T}}_4$ is Constant  $\Rightarrow \mathrm{n}-6=0$

                                                      $\mathrm{n}=6$

$\begin{array}{rcl}{}^6{\mathrm{C}}_3 {\left(\mathrm{a}\right)}^3& =& \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 20 {\mathrm{a}}^3& =& \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \mathrm{a}& =& \frac{1}{2}\end{array}$