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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.

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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?

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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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$$-48=\dfrac p{d^3}+\dfrac pd+pd+pd^3=\dfrac p{d^3}(1+d^2+d^4+d^6)=\dfrac{p(1+d^2)(1+d^4)}{d^3}$$

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 $u$

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