Consider the quadratic equation $$(n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0, n \in \mathbb{R}$$. Let $$\alpha$$ be the minimum value of the product of its roots and $$\beta$$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is $$\alpha$$ and the common ratio is $$\dfrac{\alpha}{\beta}$$, is :

Let $$k = n^2 - 2n + 2$$

$$k = (n - 1)^2 + 1$$

$$k \ge 1$$

$$k x^2 - 3x + k^2 = 0$$

$$\text{Product of roots} = \frac{C}{A} = \frac{k^2}{k} = k$$

Since $$k \ge 1$$, the minimum value of the product of roots is $$\alpha = 1$$

$$\text{Sum} = \frac{-B}{A} = \frac{3}{k}$$

To maximize $$\frac{3}{k}$$, we minimize the denominator $$k$$. Since the minimum value of $$k$$ is $$1$$: $$\beta = \frac{3}{1} = 3$$

$$S_6 = a \cdot \frac{1 - r^6}{1 - r}$$

$$S_6 = 1 \cdot \frac{1 - \left(\frac{1}{3}\right)^6}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{729}}{\frac{2}{3}}$$

$$S_6 = \frac{\frac{728}{729}}{\frac{2}{3}} = \frac{728}{729} \times \frac{3}{2} = \frac{364}{243}$$