In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals:

Let the geometric progression be $a,ar,a{r}^2,\dots$ where $a>0$ and $r>0$.According to the problem, any term $a_n$ is equal to the sum of the next two terms $a_{n+1}$ and $a_{n+2}$.
$$a{r}^{n-1}=a{r}^n+a{r}^{n+1}$$
Dividing both sides by $a{r}^{n-1}$ (since $a,r\ne 0$):
$$1=r+r^2$$
$$r^2+r-1=0$$
Using the quadratic formula $r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:
$$r=\frac{-1\pm \sqrt{1^2-4(1)(-1)}}{2(1)}=\frac{-1\pm \sqrt{5}}{2}$$
Since the terms are positive, the common ratio $r$ must be positive.
$$r=\frac{\sqrt{5}-1}{2}$$