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 first term of the geometric progression (G.P.) be $$a$$ and the common ratio be $$r$$. Thus the terms are $$a,\,ar,\,ar^{2},\,ar^{3},\ldots$$ with all terms >0.

The condition in the question states: each term equals the sum of the next two terms. Applying this to the first term, we get

$$a = ar + ar^{2}$$

Divide both sides by $$a \;(\neq 0)$$ to eliminate $$a$$:

$$1 = r + r^{2}$$

Re-arrange to obtain a quadratic in $$r$$:

$$r^{2} + r - 1 = 0 \quad -(1)$$

Use the quadratic formula for $$ax^{2}+bx+c=0$$, namely $$x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$. Here $$a=1,\;b=1,\;c=-1$$, so

$$r = \frac{-1 \pm \sqrt{1^{2}-4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$$

This yields two roots: $$r_{1} = \frac{-1 + \sqrt{5}}{2}, \qquad r_{2} = \frac{-1 - \sqrt{5}}{2}$$

Because every term of the G.P. is positive, the common ratio must be positive. The root $$r_{2}$$ is negative, so it is rejected. Hence

$$r = \frac{-1 + \sqrt{5}}{2} = \frac{\sqrt{5}-1}{2}$$

Therefore, the common ratio of the progression is $$\dfrac{\sqrt{5}-1}{2}$$.

Option D which is: $$\frac{1}{2}(\sqrt{5} - 1)$$