Let $\alpha , \beta$ be the roots of the equation $x^2-px+r=0$ and $\frac{\alpha }{2},2\beta$ be the roots of the equation $x^2-qx+r=0$. Then, the value of r is

The equation $x^2-px+r=01$ has roots $\alpha , \beta$ and the equation $x^2-qx+r=0$ has roots $\frac{\alpha }{2},2\beta$

$\therefore$ $r=\alpha \beta \text{ and }\alpha +\beta =\pi$

and $\frac{\alpha }{2}+2\beta =q$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\beta =\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{2q-p}{3}\\ \text{ and }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{2(2p-q)}{3}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha \beta =t=\frac{2}{9}(2q-p)(2p-q)\end{array}$