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(A) Let $r$ be the common ratio of the geometric progression. Then $\beta = \alpha r, \gamma = \alpha r^2, \delta = \alpha r^3$.From the given equatio

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$-2, -32$

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(A) Let $r$ be the common ratio of the geometric progression. Then $\beta = \alpha r, \gamma = \alpha r^2, \delta = \alpha r^3$.From the given equations:$\alpha + \beta = 1 \Rightarrow \alpha(1 + r) = 1 \dots (i)$$\alpha \beta = p \Rightarrow \alpha^2 r = p \dots (ii)$$\gamma + \delta = 4 \Rightarrow \alpha r^2(1 + r) = 4 \dots (iii)$$\gamma \delta = q \Rightarrow \alpha^2 r^5 = q \dots (iv)$Dividing $(iii)$ by $(i)$,we get $r^2 = 4$,so $r = \pm 2$.If $r = 2$,then $\alpha(1 + 2) = 1 \Rightarrow \alpha = 1/3$. Then $p = \alpha^2 r = (1/9)(2) = 2/9$ and $q = \alpha^2 r^5 = (1/9)(32) = 32/9$.If $r = -2$,then $\alpha(1 - 2) = 1$ $\Rightarrow -\alpha = 1$ $\Rightarrow \alpha = -1$.Then $p = \alpha^2 r = (-1)^2(-2) = -2$ and $q = \alpha^2 r^5 = (-1)^2(-2)^5 = -32$.Thus,$(p, q) = (-2, -32)$.

If $\alpha, \beta$ are the roots of $x^2 - x + p = 0$ and $\gamma, \delta$ are the roots of $x^2 - 4x + q = 0$,and if $\alpha, \beta, \gamma, \delta$ are in a geometric progression,then the values of $p$ and $q$ are respectively:

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