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