Question

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

• A. $\frac{\sqrt{3}-1}{2}<r<\frac{\sqrt{3}+1}{2}$
• B. $\frac{\sqrt{5}-1}{2}<r<\frac{\sqrt{5}+1}{2}$
• C. $\frac{\sqrt{2}-1}{2}<r<\frac{\sqrt{2}+1}{2}$
• D. None of these

Answer 1

Answer: (B)

Let the sides of the triangle be $a,ar,a{r}^2$.
If $r=1$, then the three terms of G.P. will be $a,a,a$ and hence an equilateral triangle will be formed.
Thus when $r=1$, triangle will be formed
If $r>1$, then greatest side will be $a{r}^2$ and in this case triangle will be formed if
$a+ar>a{r}^2$
$\Rightarrow r^2-r-1<0$
$\Rightarrow \frac{1-\sqrt{5}}{2}<r<\frac{1+\sqrt{5}}{2}$
$\Rightarrow r<\frac{1+\sqrt{5}}{2}$
$[\because r>1](2)$
If $r<1$, then greatest side will be $a$ and triangle will be formed if
$ar+a{r}^2>a$
$\Rightarrow r^2+r-1>0$
$\Rightarrow r<\frac{-1-\sqrt{5}}{2}$
or $r>\frac{-1+\sqrt{5}}{2}$
$\Rightarrow \frac{\sqrt{5}-1}{2}<r<1$
$[\because 0<r<1](3)$
From (1), (2) and (3), possible values of $r$ are given by
$\frac{\sqrt{5}-1}{2}<r<\frac{\sqrt{5}+1}{2}$