If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

Let the four positive consecutive terms of the G.P. be $$a,\; ar,\; ar^{2},\; ar^{3}$$ where $$a \gt 0$$ and $$r \gt 0$$ is the common ratio.

Sum of the four terms:
$$a + ar + ar^{2} + ar^{3} = 126$$
$$\Rightarrow a\,(1+r+r^{2}+r^{3}) = 126$$ $$-(1)$$

Product of the four terms:
$$a \cdot ar \cdot ar^{2} \cdot ar^{3} = 1296$$
$$\Rightarrow a^{4} r^{6} = 1296$$
$$\Rightarrow a^{4} = 1296\, r^{-6}$$
$$\Rightarrow a = 1296^{1/4}\, r^{-3/2} = 6\, r^{-3/2}$$ $$-(2)$$

Substituting $$a$$ from $$(2)$$ into $$(1)$$:

$$6\, r^{-3/2}\bigl(1 + r + r^{2} + r^{3}\bigr) = 126$$
$$\Rightarrow 1 + r + r^{2} + r^{3} = 21\, r^{3/2}$$ $$-(3)$$

Introduce the substitution $$r = x^{2}$$ with $$x \gt 0$$. Then $$r^{3/2} = x^{3}$$. Equation $$(3)$$ becomes

$$x^{6} + x^{4} + x^{2} + 1 = 21\,x^{3}$$
$$\Rightarrow x^{6} + x^{4} + x^{2} - 21x^{3} + 1 = 0$$ $$-(4)$$

Divide $$(4)$$ by $$x^{3}$$ (since $$x \gt 0$$):
$$x^{3} + x + \frac{1}{x} + \frac{1}{x^{3}} - 21 = 0$$

Define $$y = x + \dfrac{1}{x}\;(y \ge 2)$$. Using the identity $$x^{3} + \dfrac{1}{x^{3}} = y^{3} - 3y$$, the above equation becomes

$$(y^{3} - 3y) + y - 21 = 0$$
$$\Rightarrow y^{3} - 2y - 21 = 0$$ $$-(5)$$

Check integral values for $$y$$. For $$y = 3: \; 3^{3} - 2(3) - 21 = 27 - 6 - 21 = 0$$, hence $$y = 3$$ is a root.

Factorising $$(5)$$: $$(y-3)(y^{2} + 3y + 7) = 0$$

The quadratic $$y^{2} + 3y + 7 = 0$$ has negative discriminant, so the only real root is $$y = 3$$.

Thus $$x + \dfrac{1}{x} = 3$$.
Solve for $$x$$: $$x^{2} - 3x + 1 = 0$$
$$x = \dfrac{3 \pm \sqrt{9 - 4}}{2} = \dfrac{3 \pm \sqrt{5}}{2}$$

Both roots are positive, hence

Case 1: $$x_{1} = \dfrac{3 + \sqrt{5}}{2} \;\Longrightarrow\; r_{1} = x_{1}^{2} = \dfrac{(3 + \sqrt{5})^{2}}{4} = \dfrac{7 + 3\sqrt{5}}{2}$$

Case 2: $$x_{2} = \dfrac{3 - \sqrt{5}}{2} \;\Longrightarrow\; r_{2} = x_{2}^{2} = \dfrac{(3 - \sqrt{5})^{2}}{4} = \dfrac{7 - 3\sqrt{5}}{2}$$

Therefore, the common ratio can take two positive values $$r_{1}$$ and $$r_{2}$$. The required sum of these ratios is

$$r_{1} + r_{2} = \dfrac{7 + 3\sqrt{5}}{2} + \dfrac{7 - 3\sqrt{5}}{2} = \dfrac{14}{2} = 7$$

Hence, the sum of the common ratios of all such G.P.s is $$7$$. So, the correct option is Option A.