If the fourth term in the expansion of $(px+1/x{)}^n$ is $5/2$ then the value of p is?

1. $1$
2. $1/2$
3. $6$
4. $2$

${(px+\frac{1}{x})}^n={}^n{C}_0{\left(\frac{1}{x}\right)}^0\left(px{)}^{n-0}{+}^n{C}_1{\left(\frac{1}{x}\right)}^1\right(px{)}^{n-1}$

$+...{+}^n{C}_n{\left(\frac{1}{x}\right)}^n(px{)}^{n-1}$

so, the fourth in the expansion is,

${}^n{C}_3{\left(\frac{1}{x}\right)}^3(px{)}^{n-3}$

But given the fourth term is $\frac{5}{2}$ which is independent

of x. So, $n-3-3=0$

$\Rightarrow n=6$

Now, ${}^6{C}_3.P^3=\frac{5}{2}$

$\Rightarrow 20P^3=\frac{5}{2}\Rightarrow p^3=\frac{1}{8}$

$\Rightarrow p=\frac{1}{2}$