jmc

Let the roots of the degree 5 polynomial be $\frac{a}{r^2},$ $\frac{a}{r},$ $a,$ $ar,$ and $a{r}^2.$ Then by Vieta's formulas, $$\frac{a}{r^2}+\frac{a}{r}+a+ar+a{r}^2=40,$$so $$a\left(\frac{1}{r^2}+\frac{1}{r}+1+r+r^2\right)=40.(\ast )$$Also, $$\frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}=10,$$so $$\frac{1}{a}\left(r^2+r+1+\frac{1}{r}+\frac{1}{r^2}\right)=10.(\ast \ast )$$Dividing equations $(\ast )$ and $(\ast \ast ),$ we get $a^2=4,$ so $a=\pm 2.$

Again by Vieta's formulas, $$S=-\frac{a}{r^2}\cdot \frac{a}{r}\cdot a\cdot ar\cdot a{r}^2=-a^5$$so $|S|=|a^5|=|a{|}^5=\boxed{32}.$