(d)
Here, $\alpha + \beta = - \frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$
But $\alpha\beta = 3\ \ \ \Rightarrow \ \ \ \ c = 3a$
Also, $b = \frac{a + c}{2}\ \ \ \ \Rightarrow \ \ \ \ b = \frac{a + 3a}{2} = 2a$
Hence, $\alpha + \beta = - \frac{2a}{a} = - 2$