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${\int }_0^n\left\{x\right\}dx,{\int }_0^n\left[x\right]dx$ 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

Determine the value of $n$

On solving the fractional part of an integer, we get:

$\begin{array}{rcl}{\int }_0^n\left\{x\right\}dx& =& n{\int }_0^1\left\{x\right\}dx\left[\because \left\{x\right\}\mathrm{is}a\mathrm{periodic}\mathrm{function}\mathrm{with}\mathrm{period}1\right]\\ & =& n{\int }_0^{1}xdx\\ & =& \frac{n}{2}\\ & & \end{array}$

For the greatest integer:

$\begin{array}{rcl}{\int }_0^n\left[x\right]dx& =& {\int }_0^n\left(x-\left\{x\right\}\right)dx\mathbf{}\left[\because x=\left\{x\right\}+\left[x\right]\right]\\ & =& \frac{n^2}{2}-\frac{n}{2}\end{array}$

Therefore, we know that in a GP $b^2=ac$

$$\begin{gathered} {\left(\begin{array}{rcl}& & \frac{n^2}{2}-\frac{n}{2}\end{array}\right)}^2=\frac{n}{2}\left[10\left(n^2-n\right)\right] \\ \Rightarrow \left(\begin{array}{l}\frac{n^2-n}{2}\end{array}\right)=10n \\ \Rightarrow n-1=20 \\ \Rightarrow n=21\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{n}\mathbf{>}\mathbf{1}\mathbf{]} \end{gathered}$$

Hence, $n$is equal to $21$