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(C) Since $a, b$ and $c$ form a geometric progression,we have $b = ar$ and $c = ar^2$. Substituting these into the line equation $ax + by + c = 0$,we

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$\frac{r}{2}$

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(C) Since $a, b$ and $c$ form a geometric progression,we have $b = ar$ and $c = ar^2$. Substituting these into the line equation $ax + by + c = 0$,we get $ax + ary + ar^2 = 0$. Dividing by $a$ (assuming $a 
eq 0$),we get $x + ry + r^2 = 0$,which implies $x = -ry - r^2$. Substituting $x = -ry - r^2$ into the curve equation $x + 2y^2 = 0$,we get $-ry - r^2 + 2y^2 = 0$,or $2y^2 - ry - r^2 = 0$. This is a quadratic equation in $y$ of the form $Ay^2 + By + C = 0$,where $A = 2, B = -r, C = -r^2$. The sum of the ordinates (roots of the quadratic equation) is given by $-\frac{B}{A} = -\frac{-r}{2} = \frac{r}{2}$.

If $a, b$ and $c$ form a geometric progression with common ratio $r$,then the sum of the ordinates of the points of intersection of the line $ax + by + c = 0$ and the curve $x + 2y^2 = 0$ is

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