<s>[INST] Geometric sequence problem including sum of the numbers Numbers: $a,b,c,d$ generate geometric sequence and $a+b+c+d=-40. $ Find these numbers if $a^2+b^2+c^2+d^2=3280$
I tried this problem and I have system of equations which I can't solve. I think there should be different way to handle this problem.
 [/INST] Let $r$ be the common difference of the GP. Then using GP sum formula, we have,
\begin{align}
a+b+c+d=-40&\implies a\left(\dfrac{r^4-1}{r-1}\right)=-40\tag1\\
a^2+b^2+c^2+d^2=3280&\implies a^2\left(\dfrac{r^8-1}{r^2-1}\right)=3280\tag2
\end{align}
Now, divide the second equation from the square of the first equation to get,
\begin{equation}
\dfrac{(r^4+1)(r-1)}{(r^4-1)(r+1)}=\dfrac{41}{20}\tag3
\end{equation}
Solve to get,
\begin{equation}
r\in\left\{-3,-\dfrac13\right\}
\end{equation}
Now make two cases for $r$ and substitute in the first equation to get the desired answer.
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