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.
 A: Let $\frac{b}{a}=q$ and $1+q^2=2uq.$
Thus, $|u|\geq1$, $$a=-\frac{40}{1+q+q^2+q^3}$$ and $$a^2(1+q^2+q^4+q^6)=3280,$$ which gives
$$\frac{1600}{(1+q)^2(1+q^2)^2}\cdot(1+q^2)(1+q^4)=3280$$ or
$$20(1+q^4)=41(1+q)^2(1+q^2)$$ or
$$10(2u^2-1)=41(u+1)u$$ or
$$21u^2+41u+10=0,$$ which gives $$u=-\frac{5}{3},$$
$$1+q^2=-\frac{10}{3}q$$ and $$q\in\left\{-3,-\frac{1}{3}\right\}.$$
Can you end it now?
A: 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.
A: $$-48=\dfrac p{d^3}+\dfrac pd+pd+pd^3=\dfrac p{d^3}(1+d^2+d^4+d^6)=\dfrac{p(1+d^2)(1+d^4)}{d^3}$$
Similarly,
$$3280=\dfrac{p^2(1+d^4)(1+d^8)}{d^6}$$
$$\implies\dfrac{(-48)^2}{3280}=\dfrac{(1+d^2)^2(1+d^4)}{1+d^8}=\dfrac{\left(d+\dfrac1d\right)^2\left(d^2+\dfrac1{d^2}\right)}{d^4+\dfrac1{d^4}}$$
Let $d^2+\dfrac1{d^2}=u$
$$d^4+\dfrac1{d^4}=u^2-2, \left(d+\dfrac1d\right)^2=u+2$$
So, we have a quadratic equation in $u$
