Let α, β be the roots of the equation x^2 - x + p = 0 and γ, δ be the roots the equation x^2 - 4x + q = 0; p, q ∈ Z. If α, β, γ, δ are in G.P., then…

Let $\alpha = a, \beta = ar, \gamma = ar^2, \delta = ar^3$α=a,β=ar,γ=ar2,δ=ar3 be the terms of the G.P.

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

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

$$\gamma + \delta = 4 \Rightarrow ar^2(1 + r) = 4$$γ+δ=4⇒ar2(1+r)=4

Dividing the second equation by the first equation:

$$\dfrac{ar^2(1 + r)}{a(1 + r)} = \dfrac{4}{1}$$a(1+r)ar2(1+r)​=14​

$r^2 = 4 \Rightarrow r = 2$r2=4⇒r=2 or $r = -2$r=−2

If $r = 2$r=2, then $a(1 + 2) = 1 \Rightarrow a = \dfrac{1}{3}$a(1+2)=1⇒a=31​.

The product of the roots of the first equation is $p = \alpha \beta = a^2 r = \left(\dfrac{1}{3}\right)^2 \times 2 = \dfrac{2}{9}$p=αβ=a2r=(31​)2×2=92​. Since $p \in \mathbf{Z}$p∈Z, $r = 2$r=2 is rejected.

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

The product of the roots of the first equation is $p = \alpha \beta = a^2 r = (-1)^2 \times (-2) = -2$p=αβ=a2r=(−1)2×(−2)=−2.

The product of the roots of the second equation is $q = \gamma \delta = (ar^2)(ar^3) = a^2 r^5 = (-1)^2 \times (-2)^5 = -32$q=γδ=(ar2)(ar3)=a2r5=(−1)2×(−2)5=−32.

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

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

Answer: $34$34