Suppose the sides of a triangle form a geometric progression with common ratio $r$. Then, $r$ lies in the interval
$\left(0, \frac{-1+\sqrt{5}}{2}\right)$
$\left(\frac{1+\sqrt{5}}{2}, \frac{2+\sqrt{5}}{2}\right)$
$\left(\frac{1+\sqrt{5}}{2}, \frac{2+\sqrt{5}}{2}\right]$
$\left(\frac{2+\sqrt{5}}{2}, \infty\right)$
If the product of three consecutive terms of $G.P.$ is $216$ and the sum of product of pair-wise is $156$, then the numbers will be
The sum of two numbers is $6$ times their geometric mean, show that numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$
The number which should be added to the numbers $2, 14, 62$ so that the resulting numbers may be in $G.P.$, is
if $x = \,\frac{4}{3}\, - \,\frac{{4x}}{9}\, + \,\,\frac{{4{x^2}}}{{27}}\, - \,\,.....\,\infty $ , then $x$ is equal to
A $G.P.$ consists of an even number of terms. If the sum of all the terms is $5$ times the sum of terms occupying odd places, then find its common ratio.