(c)

Let the sides of triangle are

$a, a r, a r^2 . \quad[\because$ sides of triangle in $GP ]$

Case I $r > 1$

We know sum of two sides is greater than third side.

$\therefore a+a r>a r^2 \Rightarrow r^2-r-1<0$

$\begin{array}{ll} \Rightarrow & r=\frac{1 \pm \sqrt{5}}{2} \\\Rightarrow & 1 < r < \frac{\sqrt{5}+1}{2}, r > 1\end{array}$

Case $II$ $0 < r < 1$

$\therefore a r^2+a r > a \Rightarrow r^2+r-1 > 0$

$\Rightarrow \quad r=\frac{-1 \pm \sqrt{5}}{2}$

$\Rightarrow \quad 1 > r > \frac{\sqrt{5}-1}{2}, 0 < r < 1$

$\therefore \quad r \in\left(\frac{\sqrt{5}-1, \sqrt{5}+1}{2}, \frac{2}{2}\right)$