Question

Suppose the sides of a triangle form a geometric progression with common ratio $r$. Then, $r$ lies in the interval

• A. $\left(0, \frac{-1 +\sqrt{5}}{2}\right)$
• B. $\left(\frac{1+\sqrt{5}}{2},\frac{2+\sqrt{5}}{2}\right)$
• C. $\left(\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$
• D. $\left(\frac{2+\sqrt{5}}{2}, \infty\right)$

Answer 1

Answer: (C)

Let the sides of triangle are
$a, ar, ar^{2}$
$[\because$ sides of triangle in $GP]$
Case I $r >\,1$
We know sum of two sides is greater than third side
$\therefore a+ar >\,ar^{2}$
$\Rightarrow r^{2}-r-1 <\,0$
$\Rightarrow r=\frac{1\pm\sqrt{5}}{2}$
$\Rightarrow 1 <\,r <\frac{\sqrt{5}+1}{2}, r >\,1$
Case II $0 <\, r <\,1$
$\therefore ar^{2}+ar >\,a$
$\Rightarrow r^{2}+r-1>\,0$
$\Rightarrow r=\frac{-1 \pm\sqrt{5}}{2}$
$\Rightarrow 1>\,r >\frac{\sqrt{5}-1}{2}, 0<\,r <\,1$
$\therefore r \in\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$