Let \(\alpha, \beta\) be the roots of the equation \(x^2-x+p=0\) and \(\gamma, \delta\) be the roots of the equation \(x…

Let \(\alpha, \beta, \gamma, \delta\) be \(a, a r, a r^2, a r^3\), respectively.
For \(x^2-x+p=0\),

sum of roots \(=1\),

so \(a+a r=a(1+r)=1\)

For \(x^2-4 x+q=0\),

sum of roots \(=4\),

so \(a r^2+a r^3=a r^2(1+r)=4\)

Dividing,

\(\frac{a r^2(1+r)}{a(1+r)}=\frac{4}{1}\)

\(r^2=4\)

So, \(r=2\) or \(r=-2\).
If \(r=2\), then \(a(3)=1 \Rightarrow a=\frac{1}{3}\),

so \(p=\alpha \beta=a \cdot a r=a^2 r=\frac{2}{9}\), not an integer.

Hence \(r=-2\).
Now \(a(1-2)=1 \Rightarrow-a=1 \Rightarrow a=-1\).
So,

\(p=\alpha \beta=a \cdot a r=a^2 r=-2\)

Also,

\(q=\gamma \delta=a r^2 \cdot a r^3=a^2 r^5=(-2)^5=-32\)

\(|p+q|=|-2-32|=34\)