If the roots of the equation $x^4 + ax^3 + bx^2 + cx + d = 0$ are in geometric progression then, If the roots of the equation $x^4 + ax^3 + bx^2 + cx + d = 0$ are in geometric progression then,
$a) b^2 = ac$
$b) a^2 = b$
$c) c^2 = a^2d$
Using Vieta's relations, finding values of coefficients in terms of the assumed roots is what first comes to mind. That process works out well to give the answer ($c)$) But, as it can be seen, this method is quite lengthy. Is there a shorter method to solve this question or a trick?
 A: Here is the OP's original method. Assume the roots are $k, kr, kr^2, kr^3$. Then:
$$k(1+r+r^2+r^3) = -a$$
$$k^2r+k^2r^3+k^2r^4+k^2r^3+k^2r^4+k^3r^5 = b$$
$$k^3r^3 + k^3r^4+k^3r^5+k^3r^6 = k^3 r^3(1+r+r^2+r^3) = -c$$
$$k^4r^6 = d$$
By comparing the maximum powers of $r$, a) is not correct ($r^{10}$ vs $r^9$) and so is b). But $c/a = k^2r^3$ or $c^2/a^2 = k^4r^6 = d$, hence c) is correct.
This relationship between $a$ and $c$ can be extended to any power $n$ in general. $-c$ can also be written as $k^4 r^6 (1/k + 1/(kr) + 1/(kr^2) + 1/(kr^3))$ as you miss out one term in the product of three (or $n$) roots, and in general this is $k^n r^{n(n-1)/2} (1/k + \cdots + \frac{1}{kr^{n-1}}) = k^{n-1} r^{(n-1)(n-2)/2} (1 + \cdots + r^{n-1})$.
A: Hint: let the roots be $t,u,v,w$, then a sufficient condition for a GP is $tw=uv$ $\implies tw=uv=\sqrt{tuvw}=\sqrt{d}$.
This requires $d\gt 0$, and in that case let $x=x'\sqrt[4]{d}$ then the polynomial
$$dx'^4+a\sqrt[4]{d^3} x'^3+b\sqrt{d} x'^2+c\sqrt[4]{d} x' +d $$
has roots $t'=t/\sqrt[4]{d}, u'=u/\sqrt[4]{d},\dots$ with $t'w'=u'v'=1$ i.e. it is reciprocal, so $a\sqrt[4]{d^3}=c\sqrt[4]{d}\,$.
