If $a$ and $b$ are the roots of $x^{2}-3x+p=0$ and $c$ and $d$ are roots of $x^{2}-12x+q=0$,where $a, b, c, d$ form a $G.P.$,prove that $(q+p):(q-p)=17:15$.

Given that $a$ and $b$ are the roots of $x^{2}-3x+p=0$,we have $a+b=3$ and $ab=p$ $(1)$.
Given that $c$ and $d$ are the roots of $x^{2}-12x+q=0$,we have $c+d=12$ and $cd=q$ $(2)$.
Since $a, b, c, d$ are in $G.P.$,let $a=x, b=xr, c=xr^{2}, d=xr^{3}$.
From $(1)$,$x+xr=3 \Rightarrow x(1+r)=3$.
From $(2)$,$xr^{2}+xr^{3}=12 \Rightarrow xr^{2}(1+r)=12$.
Dividing the two equations: $\frac{xr^{2}(1+r)}{x(1+r)} = \frac{12}{3}$ $\Rightarrow r^{2}=4$ $\Rightarrow r=\pm 2$.
Case $I$: If $r=2$,then $x(1+2)=3 \Rightarrow x=1$. Thus $a=1, b=2, c=4, d=8$. Then $p=ab=2$ and $q=cd=32$.
$\frac{q+p}{q-p} = \frac{32+2}{32-2} = \frac{34}{30} = \frac{17}{15}$.
Case $II$: If $r=-2$,then $x(1-2)=3 \Rightarrow x=-3$. Thus $a=-3, b=6, c=-12, d=24$. Then $p=ab=-18$ and $q=cd=-288$.
$\frac{q+p}{q-p} = \frac{-288-18}{-288+18} = \frac{-306}{-270} = \frac{17}{15}$.
Thus,in both cases,$(q+p):(q-p)=17:15$.