Question:

From $1, \alpha + \beta, \alpha\beta$ in A.P.: $1 + \alpha\beta = 2(\alpha + \beta)$ gives $1 + \alpha\beta = \alpha + \alpha + 2b = 0$ (equation 1). From $1, \frac{1}{\alpha}, \frac{1}{\beta}$ in A.P.: $\frac{2}{\alpha} = 1 + \frac{1}{\beta}$, which simplifies to $a = -b = c$ (equation 2). Since $\alpha, \beta$ are roots of $x^2 - x + 1 = 0$, we compute $\frac{\alpha^2 + \beta^2}{2\alpha^2\beta^2} = \frac{1}{2}(\alpha^2 + \beta^2) - \frac{1}{2} + 1 = 1.5$.