Question

Let $\left\{x\right\}$ and $\left[x\right]$ denote the fractional part of $x$ and the greatest integer $\le x$ respectively of a real number $x.$ If $\underset{0}{\overset{n}{\int }}\left\{x\right\}dx,\underset{0}{\overset{n}{\int }}\left[x\right]dx$ and $10(n^2-n),(n\in N,n>1)$ are three consecutive terms of a G.P., then $n$ is equal to

Answer 1

$\underset{0}{\overset{n}{\int }}\left\{x\right\}dx=n\underset{0}{\overset{1}{\int }}xdx$
$n{\left(\frac{x^2}{2}\right)}_0^1$
$\therefore \underset{0}{\overset{n}{\int }}\left\{x\right\}dx=\frac{n}{2}$

$\underset{0}{\overset{n}{\int }}\left[x\right]dx=\underset{0}{\overset{1}{\int }}0dx+\underset{1}{\overset{2}{\int }}1dx+\underset{2}{\overset{3}{\int }}2dx+\dots +\underset{n-1}{\overset{n}{\int }}(n-1)dx$
$=1+2+3+\dots +(n-1)$
$\therefore \underset{0}{\overset{n}{\int }}\left[x\right]dx=\frac{n(n-1)}{2}$

$\frac{n}{2},\frac{n(n-1)}{2},10(n^2-n)$ are in G.P.
${\left(\frac{n(n-1)}{2}\right)}^2=\frac{n}{2}\cdot 10(n^2-n)$
$\Rightarrow n-1=20$
$\therefore n=21$