Let $$\alpha$$ and $$\beta$$ be the roots of the equation $$px^2 + qx - r = 0$$, where $$p \neq 0$$. If $$p$$, $$q$$ and $$r$$ be the consecutive terms of a non-constant G.P and $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}$$, then the value of $$(\alpha - \beta)^2$$ is:

We have $$\alpha, \beta$$ as roots of $$px^2 + qx - r = 0$$ where $$p \neq 0$$. The terms $$p, q, r$$ form a non-constant G.P., and $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}$$.

From $$px^2 + qx - r = 0$$:

• $$\alpha + \beta = -\frac{q}{p}$$
• $$\alpha\beta = -\frac{r}{p}$$

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-q/p}{-r/p} = \frac{q}{r} = \frac{3}{4}$$

Since $$p, q, r$$ are consecutive terms of a G.P. with common ratio $$k$$: $$q = pk$$ and $$r = pk^2$$.

$$\frac{q}{r} = \frac{pk}{pk^2} = \frac{1}{k} = \frac{3}{4} \implies k = \frac{4}{3}$$

So $$q = \frac{4p}{3}$$ and $$r = \frac{16p}{9}$$.

$$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$$

$$= \left(\frac{q}{p}\right)^2 + \frac{4r}{p} = \left(\frac{4}{3}\right)^2 + 4 \times \frac{16}{9}$$

$$= \frac{16}{9} + \frac{64}{9} = \frac{80}{9}$$

The correct answer is Option A: $$\frac{80}{9}$$.