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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Solution
a, ar, ar² → G.P.
Sum of any two sides $>$ third side
$$
\begin{aligned}
& a+a r>a r^2, a+a r^2>a r, a r+a r^2>a \\
& r^2-r-1<0 \\
& r \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right) \\
& r^2-r+1>0
\end{aligned}
$$
always true
$$
\begin{aligned}
& r^2+r-1>0 \\
& r \in\left(-\infty,-\frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)
\end{aligned}
$$
Taking intersection of (1), (2)
$$
\begin{aligned}
& r \in\left(\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right) \\
& \text { As } r>1 \\
& r \in\left(1, \frac{1+\sqrt{5}}{2}\right) \\
& {[r]=1[-r]=-2} \\
& 3[r]+[-r]=1
\end{aligned}
$$
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