Let x and [x] denote the fractional part of x and the greatest integer ≤ x respectively of a real number x. if ∫ _0^n xdx,∫ _0^n[x]dx and…

$${\int }_{0}^{n}{x}dx=n{\int }_{0}^{1}xdx=n{(\frac{{x}^{2}}{2})}_{0}^{1}=\frac{n}{2}$$∫0n​xdx=n∫01​xdx=n(2x2​)01​=2n​

and $\displaystyle {\int }_{0}^{n}[x]dx={\int }_{0}^{n}(x-{x})dx={(\frac{{x}^{2}}{2})}_{0}^{n}-{\int }_{0}^{n}{x}dx=\frac{{n}^{2}}{2}-\frac{n}{2}$∫0n​[x]dx=∫0n​(x−x)dx=(2x2​)0n​−∫0n​xdx=2n2​−2n​

Now, $\frac{n}{2},\frac{{n}^{2}-n}{2}$2n​,2n2−n​ and $10({n}^{2}-n)$10(n2−n) are in Geometric progression

So,${(\frac{{n}^{2}-n}{2})}^{2}=\frac{n}{2}\cdot 10({n}^{2}-n)$(2n2−n​)2=2n​⋅10(n2−n)

$$\Rightarrow \frac{{n}^{2}(n-1{)}^{2}}{4}=5.{n}^{2}(n-1)$$⇒4n2(n−1)2​=5.n2(n−1)

$\Rightarrow n-1=20\Rightarrow n=21$⇒n−1=20⇒n=21 $(\because n\neq 1)$(∵n=1)