Mathematics - Sequence & Series Question with Solution | TestHub

Let the three numbers of the GP be$a, ar$and$a{r}^2$where$r$is the common ratio.
According to the given condition,$a, 2ar$and$a{r}^2$are in AP.

$\therefore 2ar=\frac{a+a{r}^2}{2}$

$\therefore a+a{r}^2=4ar$

$\therefore r^2+1=4r$

$\therefore r^2-4r+1=0$

$\therefore r=\frac{4\pm \sqrt{16-4}}{2}$

$=\frac{4\pm 2\sqrt{3}}{2}=2\pm \sqrt{3}$

As it is an 'increasing' GP,$r > 1$.

$\therefore r=2+\sqrt{3}$.