- Home
- Standard 11
- Mathematics
If three successive terms of a$G.P.$ with common ratio $r(r>1)$ are the lengths of the sides of a triangle and $[\mathrm{r}]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to :
$1$
$2$
$3$
$4$
Solution
$\text { a, ar, } a r^2 \rightarrow \text { G.P. }$
Sum of any two sides $>$ third side
$ a+a r>a r^2, a+a r^2>a r, a r+a r^2>a $
$ r^2-r-1<0 $
$ r \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right) $
$ r^2-r+1>0$ $…………(1)$
always true
$ \mathrm{r}^2+\mathrm{r}-1>0 $
$ \mathrm{r} \in\left(-\infty,-\frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)$ $……………(2)$
Taking intersection of $(1)$, $(2)$
$\mathrm{r} \in\left(\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$
As $\mathrm{r}>1$
$ r \in\left(1, \frac{1+\sqrt{5}}{2}\right) $
$ {[r]=1[-r]=-2} $
$ 3[r]+[-r]=1$
