If the $4^{\text{th}}$ term in the expansion of $\left( ax + \frac{1}{x} \right)^n$ is $\frac{5}{2}$, then the values of $a$ and $n$ are

The general term is $$T_{r+1}=\binom{n}{r}(ax)^{n-r}\left(\frac{1}{x}\right)^r.$$For the $4^{\text{th}}$ term, $r=3$. So $$T_4=\binom{n}{3}a^{n-3}x^{n-6}.$$Since the $4^{\text{th}}$ term is $\frac{5}{2}$, it is a constant term. Hence $$n-6=0.$$So $$n=6.$$Now $$T_4=\binom{6}{3}a^3=\frac{5}{2}.$$Thus $$20a^3=\frac{5}{2}.$$So $$a^3=\frac{1}{8}.$$Hence $$a=\frac{1}{2}.$$Therefore the pair is $\left(\frac{1}{2},6\right)$, which matches option $A$.