Let α, β be the roots of the equation x^2 - x + p = 0 and γ, be the...

Let $\alpha = a, \beta = ar, \gamma = ar^2, \delta = ar^3$ be the terms of the G.P.

From the given quadratic equations, the sum of the roots are:

$\alpha + \beta = 1 \Rightarrow a(1 + r) = 1$

$\gamma + \delta = 4 \Rightarrow ar^2(1 + r) = 4$

Dividing the second equation by the first equation:

$\dfrac{ar^2(1 + r)}{a(1 + r)} = \dfrac{4}{1}$

$r^2 = 4 \Rightarrow r = 2$ or $r = -2$

If $r = 2$, then $a(1 + 2) = 1 \Rightarrow a = \dfrac{1}{3}$.

The product of the roots of the first equation is $p = \alpha \beta = a^2 r = \left(\dfrac{1}{3}\right)^2 \times 2 = \dfrac{2}{9}$. Since $p \in \mathbf{Z}$, $r = 2$ is rejected.

If $r = -2$, then $a(1 - 2) = 1 \Rightarrow a = -1$.

The product of the roots of the first equation is $p = \alpha \beta = a^2 r = (-1)^2 \times (-2) = -2$.

The product of the roots of the second equation is $q = \gamma \delta = (ar^2)(ar^3) = a^2 r^5 = (-1)^2 \times (-2)^5 = -32$.

Both $p$ and $q$ are integers, which satisfies the given condition.

Therefore, $|p + q| = |-2 - 32| = |-34| = 34$.

Answer: $34$