Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is:

Let the three positive numbers in increasing geometric progression be denoted by $$\dfrac{a}{r},\;a,\;ar$$ where $$a>0$$ is the geometric mean and $$r>1$$ is the common ratio (because the progression is increasing).

The problem states that the middle term is doubled. Therefore the new three numbers become

$$\dfrac{a}{r},\;2a,\;ar.$$

These new numbers are said to form an arithmetic progression. For any three numbers $$x,\;y,\;z$$ to be in A.P., the defining condition is

$$2y = x + z.$$

Applying this condition to our sequence $$\dfrac{a}{r},\;2a,\;ar$$ gives

$$2 \times 2a \;=\; \dfrac{a}{r} \;+\; ar.$$

Simplify step by step:

$$4a \;=\; \dfrac{a}{r} + ar.$$

Now divide every term by $$a$$ (recall $$a>0$$):

$$4 \;=\; \dfrac{1}{r} + r.$$

Bring all terms to one side so that zero appears on the other side:

$$4 - r - \dfrac{1}{r} = 0.$$

To clear the fraction, multiply the entire equation by $$r$$:

$$4r - r^2 - 1 = 0.$$

Rewrite in the standard quadratic form $$r^2 - 4r + 1 = 0$$ by multiplying through by $$-1$$:

$$r^2 - 4r + 1 = 0.$$

We now solve this quadratic using the quadratic formula. For an equation $$ur^2 + vr + w = 0,$$ the roots are $$r = \dfrac{-v \pm \sqrt{v^{2} - 4uw}}{2u}.$$

Here $$u = 1,\; v = -4,\; w = 1.$$ Substituting, we obtain

$$r = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times 1}}{2 \times 1} \;=\; \dfrac{4 \pm \sqrt{16 - 4}}{2} \;=\; \dfrac{4 \pm \sqrt{12}}{2}.$$

Simplify $$\sqrt{12} = 2\sqrt{3}$$ and divide numerator and denominator by $$2$$:

$$r = 2 \pm \sqrt{3}.$$

We must choose the value of $$r$$ that preserves the fact that the original G.P. is increasing, i.e.\ $$r > 1.$$

Compute each possibility:

$$2 + \sqrt{3} > 1 \quad\text{(true)},$$

$$2 - \sqrt{3} \approx 2 - 1.732 = 0.268 < 1 \quad\text{(false for an increasing sequence)}.$$

Therefore, the only admissible common ratio is

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

Hence, the correct answer is Option B.