If three successive terms of aG.P. with commo | vedclass

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Question

$	ext { a, ar, } a r^2 ightarrow 	ext { 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 $

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$	ext { a, ar, } a r^2 ightarrow 	ext { 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}ight) $

$ 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}ight) \cup\left(\frac{-1+\sqrt{5}}{2}, \inftyight)$               $...............(2)$

Taking intersection of $(1)$, $(2)$

$\mathrm{r} \in\left(\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}ight)$

As $\mathrm{r}>1$

$ r \in\left(1, \frac{1+\sqrt{5}}{2}ight) $

$ {[r]=1[-r]=-2} $

$ 3[r]+[-r]=1$

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 :

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