<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 $\frac{b}{a}=q$ 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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