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(C) Let the sides of the triangle be $a, ar, ar^2$ where $a > 0$ and $r > 0$. By the triangle inequality,the sum of any two sides must be greater than

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

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(C) Let the sides of the triangle be $a, ar, ar^2$ where $a > 0$ and $r > 0$. By the triangle inequality,the sum of any two sides must be greater than the third side. $1$. $a + ar > ar^2$ $\Rightarrow 1 + r > r^2$ $\Rightarrow r^2 - r - 1 The roots of $r^2 - r - 1 = 0$ are $r = \frac{1 \pm \sqrt{5}}{2}$. Since $r > 0$,we have $r $2$. $ar + ar^2 > a \Rightarrow r^2 + r - 1 > 0$. The roots of $r^2 + r - 1 = 0$ are $r = \frac{-1 \pm \sqrt{5}}{2}$. Since $r > 0$,we have $r > \frac{\sqrt{5} - 1}{2}$. Combining these,we get $\frac{\sqrt{5} - 1}{2} < r < \frac{\sqrt{5} + 1}{2}$.

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

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