Verified answer

Step-by-step explanation:

Let the roots of the equation $x^3-40x^4+Px^3+Qx^2+Rx+S=0$ be in geometric progression. Then, the sum of their reciprocals is $10$. Find $|S|$.

We can assume the roots are $a,ar,ar^2$. Then, we have $a+ar+ar^2 = \frac{P}{40}$, $a(ar^2)+ar(a)+ar^2(a) = \frac{R}{40}$, and $a(ar) = \frac{S}{40}$. Also, $\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2} = 10$, so $\frac{a^2r+a+1}{a^2r^2}=10$. This simplifies to $a^2r^3 - 10a^2r^2 + a^2r - 10 = 0.$

Using Vieta's, we get $a+ar+ar^2 = \frac{P}{40} = \frac{10}{a}$, which leads to $a^2r^2+a+1 = 0$ and $a(ar^2)+ar(a)+ar^2(a) = \frac{R}{40} = \frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}=10$. This means that $a+ar+ar^2 = 10a^2r^2$.

We then use $(a+ar+ar^2)^2 = a^2r^4 + 2a^2r^3 + 3a^2r^2 + 2a^2r + a^2 = 100a^4r^4$ and $a^2r^2+a+1=0$ to get $2a^2r^3+2a^2r+1=0$. Therefore, we have $10a^2r^2 = (a+ar+ar^2)^2 = a^2+2a^2r^2+2a^2r+2a^2r^3+2a^3r^2+2a^3r^3 = 3a^2+4a^2r+2(2a^2r^3+2a^3r^3) = 3a^2+4a^2r-4,$ so $a^2r = \frac{7}{2}$.

Finally, we have $a(ar^2)+ar(a)+ar^2(a) = \frac{S}{40} = \frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}=10,$ so $S = 4000$. Therefore, $|S|=\boxed{4000}$.