Three successive terms of a G.P. will form the sides of a triangle if the common ratio $r$satisfies the inequality

Let the lengths of the sides of the triangle be $a,ar,a{r}^2$ We have the following three cases :

CASE I    When $r=1$

In this case, the lengths of the sides of the triangle are $a, a, a$ i.e. the triangle is equilateral.

CASE II When $r>1$

In this case, the length of the largest side is $a{r}^2$ Therefore, the triangle will be formed, if

 $\begin{array}{l}a+ar>a{r}^2\\ \Rightarrow r^2-r-1<0\\ \Rightarrow \frac{1-\sqrt{5}}{2}<r<\frac{1+\sqrt{5}}{2}\mid \\ \Rightarrow r<\frac{1+\sqrt{5}}{2} [\because r>1] ..\left(i\right)\\ \end{array}$

CASE III When $r<1$

In this case, the length of the largest side is $a$. So, the triangle will be formed, if

$\begin{array}{l}ar+a{r}^2>a\\ \Rightarrow r^2+r-1>0\\ \Rightarrow r<\frac{-1}{2}-\sqrt{5 } \text{ or },r>\frac{-1+\sqrt{5}}{2}\\ \end{array}$

$\begin{array}{l}\Rightarrow \frac{\sqrt{5}-1}{2}<r<1[\because r<1]\end{array}$                         …(ii)

from (i) and (ii) we obtain: $\frac{\sqrt{5}-1}{2}<r<\frac{\sqrt{5}+1}{2}$.