Roots of Quadratic Equations forming a Geometric Progression

Given that $\alpha, \beta$ are the roots of the quadratic equation $x^2 - x + p = 0$ and $\gamma, \delta$ are the roots of $x^2 - 4x + q = 0$, where $p, q \in \mathbb{Z}$.

By Vieta's formulas, we have: $\alpha + \beta = 1 \quad \text{and} \quad \alpha \beta = p$ $\gamma + \delta = 4 \quad \text{and} \quad \gamma \delta = q$

Since $\alpha, \beta, \gamma, \delta$ are in Geometric Progression (G.P.), let the terms be represented as: $\alpha = a, \quad \beta = ar, \quad \gamma = ar^2, \quad \delta = ar^3$ where $a$ is the first term and $r$ is the common ratio.

Substituting these terms into the equations for the sum of roots: $\alpha + \beta = a + ar = a(1 + r) = 1 \quad \text{--- (1)}$ $\gamma + \delta = ar^2 + ar^3 = ar^2(1 + r) = 4 \quad \text{--- (2)}$

Dividing equation (2) by equation (1): $\frac{ar^2(1 + r)}{a(1 + r)} = \frac{4}{1}$ $r^2 = 4 \implies r = \pm 2$

Now, we evaluate the two possible cases for $r$ using the condition $p, q \in \mathbb{Z}$:

Case 1: $r = 2$ From equation (1): $a(1 + 2) = 1 \implies a = \frac{1}{3}$

Using the product of roots: $p = \alpha \beta = a^2 r = \left(\frac{1}{3}\right)^2 (2) = \frac{2}{9}$ Since $p = \frac{2}{9} \notin \mathbb{Z}$, this case is rejected.

Case 2: $r = -2$ From equation (1): $a(1 - 2) = 1 \implies a = -1$

Using the products of roots: $p = \alpha \beta = a^2 r = (-1)^2 (-2) = -2 \in \mathbb{Z}$ $q = \gamma \delta = a^2 r^5 = (-1)^2 (-2)^5 = -32 \in \mathbb{Z}$

Since both $p$ and $q$ are integers, $r = -2$ is the valid ratio.

Now, calculating $|p + q|$: $|p + q| = |(-2) + (-32)| = |-34| = 34$

Hence, the correct option is C.