Let x and x denote the fractional part of x and the greatest integer ≤ x respectively of a real number…

${\int }_0^n\left\{x\right\}\mathrm{d}x=n{\int }_0^{1}xdx=n{\left(\frac{x^2}{2}\right)}_0^1=\frac{n}{2}$
and ${\int }_0^n\left[x\right]\mathrm{d}x={\int }_0^n\left(x-\left\{x\right\}\right)\mathrm{d}x={\left(\frac{x^2}{2}\right)}_0^n-{\int }_0^n\left\{x\right\}dx=\frac{n^2}{2}-\frac{n}{2}$
Now, $\frac{n}{2},\frac{n^2-n}{2}$ and $10\left(n^2-n\right)$ are in Geometric progression

So,${\left(\frac{n^2-n}{2}\right)}^2=\frac{n}{2}·10\left(n^2-n\right)$
$\Rightarrow \frac{n^2(n-1{)}^2}{4}=5.n^2(n-1)$
$\Rightarrow n-1=20\Rightarrow n=21$ $\left(\because n\ne 1\right)$