Let $$\alpha$$, $$\beta$$ be the roots of $$x^{2} - x + p = 0$$ and $$\gamma$$, $$\delta$$ be the roots of $$x^{2}- 4x + q = 0$$ where p and q are integers. If $$\alpha, \beta, \gamma, \delta$$ are in geometric progression then p + q is

Let $$\beta=\alpha r$$, $$\gamma=\alpha r^2$$, and $$\delta=\alpha r^3$$, where r is the common difference.

$$x^{2} - x + p = 0$$

=> $$\alpha+\beta=1$$ => $$\alpha(1+r)=1\rightarrow1$$

and, $$\alpha\beta=p$$

$$x^{2}- 4x + q = 0$$

=> $$\gamma+\delta=1$$ => $$\alpha r^2(1+r)=4\rightarrow2$$

and, $$\gamma\delta=q$$

Dividing eq. 1 and eq. 2 - 

=> $$r^2=4$$ => $$r=\pm2$$

Case-1: If r = 2

=> $$\alpha(1+r)=1$$ => $$\alpha(1+2)=1$$ => $$\alpha = \dfrac{1}{3}$$

=> $$\beta=\alpha r=\dfrac{2}{3}$$

=> $$\gamma=\alpha r^2=\dfrac{4}{3}$$

=> $$\delta=\alpha r^3=\dfrac{8}{3}$$

=> $$\alpha\beta=p$$ => $$p=\left(\dfrac{1}{3}\right)\left(\dfrac{2}{3}\right)=\dfrac{2}{9}$$, but it is given that p and q are integers. Thus, this case is not valid.

Case-2: If r = -2

=> $$\alpha(1+r)=1$$ => $$\alpha(1-2)=1$$ => $$\alpha = -1$$

=> $$\beta=\alpha r=2$$

=> $$\gamma=\alpha r^2=-4$$

=> $$\delta=\alpha r^3=8$$

=> $$\alpha\beta=p$$ => $$p=(-1)(2)=-2$$

=> $$\gamma\delta=q$$ => $$q=(-4)(8)=-32$$

=> $$p+q=(-2)+(-32) = -34$$