If the middle term of ${\left(x^2+\frac{1}{x}\right)}^n$ is $924 x^{6 },$ then value of $n$ is

As there is just one middle term, $n$ must be even and 

middle term is $\left(\frac{n}{2}+1\right)$ th term, and it is given by

$\begin{array}{l}{T}_{n/2+1}{=}^n{C}_{n/2}{\left(x^2\right)}^{n-n/2}{\left(\frac{1}{x}\right)}^{n/2}\\ {=}^n{C}_{n/2}{x}^{n/2}\end{array}$

Now, ${ }^n{C}_{n/2}{x}^{n/2}=924x^6$

$\to n/2=6\to n=12$

[Also, note that  $\left.{ }^{12}{C}_6=924\right]$