Question

The three successive terms of a GP will form the sides of a triangle if the common ratio $r$ has the range $(r>1)$

• A. $(1,\frac{\sqrt{5}+1}{2})$
• B. $(-1,\frac{\sqrt{5}-1}{2})\cup (\frac{\sqrt{5}+1}{2},1)$
• C. $[1,\sqrt{5}]$
• D. $(1,1+\sqrt{5})$

Answer 1

Answer: (A)

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

Let the sides of terms of the GP be $\frac{a}{r}$ , $a$ , $ar$
Where $r$ is the common ratio and $a$ is any real number $a\ne 0$.
By applying the triangle inequality, we have
$\frac{a}{r}+a>ar$
$⟹1+r>r^2$
$⟹r^2-r-1<0$
Also, given $r>1$
Hence, acceptable solution of the quadratic is :
$r=\frac{1+\sqrt{5}}{2}$