Let $\alpha, \beta$ be the roots of the equation $x^2-x+p=0$ and $\gamma, \delta$ be the roots of $x^2-4x+q=0$. If $\alpha, \beta, \gamma, \delta$ are in G.P., then the integral values of p and q respectively, are?

Roots are in GP: a/r, a, ar, ar^2. Eq1: x^2-x+p=0 => sum=1, prod=p. Eq2: x^2-4x+q=0 => sum=4, prod=q. Let roots be a/r, a, ar, ar^2. Sums: a/r + a = 1 and ar + ar^2 = 4. a(1/r + 1) = 1 and ar(1+r) = 4. Divide: (ar(1+r)) / (a(1/r+1)) = 4/1 => r^2 = 4 => r=2 or -2. If r=2, a(1/2+1)=1 => a(3/2)=1 => a=2/3. Roots: 1/3, 2/3, 4/3, 8/3. p = (1/3)(2/3) = 2/9. q = (4/3)(8/3) = 32/9. The question asks for integral values, but these are not integers. Re-check: maybe roots are a, ar, ar^2, ar^3? Sums: a+ar=1, ar^2+ar^3=4 => a(1+r)=1, ar^2(1+r)=4 => r^2=4 => r=2 or -2. If r=-2, a(1-2)=1 => a=-1. Roots: -1, 2, -4, 8. p = (-1)*2 = -2. q = (-4)*8 = -32.