Sequences And Series Question 4
Question 4 - 2024 (01 Feb Shift 2)
If three successive terms of a G.P. with common ratio $r(r>1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to :
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Answer (1)
Solution
a, ar, $ar^{2} \rightarrow$ G.P.
Sum of any two sides $>$ third side
$a+a r>a r^{2}, a+a r^{2}>a r, a r+a r^{2}>a$
$r^{2}-r-1<0$
$r \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right) \ldots$
$r^{2}-r+1>0$
always true
$r^{2}+r-1>0$
$r \in\left(-\infty,-\frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)$.
Taking intersection of (1), (2)
$r \in\left(-\frac{1+\sqrt{ } 5}{2}, \frac{1+\sqrt{ } 5}{2}\right)$
As $r>1$
$r \in\left(1, \frac{1+\sqrt{5}}{2}\right)$
$[r]=1[-r]=-2$
$3[r]+[-r]=1$
