Question

A soma de quatro termos consecutivos de uma progressão aritmética é -6, o produto do primeiro deles pelo quarto é -54. Determinar esses termos.

Answer 1

$\mathrm{PA=(a_1,a_2,a_3,a_4,\dots)\ \to\ \boxed{\mathrm{a_1.a_4=-54}}\ \mathbf{(I)}}$

$\mathbf{Pela\ soma\ dos\ termos\ de\ uma\ PA,\ obteremos:}\\\\ \boxed{\mathbf{S_n=\dfrac{(a_1+a_n).n}{2}}}\ \to\ \mathrm{a_1+a_2+a_3+a_4=-6}\\\\ \mathrm{S_4=-6\ \to\ \dfrac{(a_1+a_4).4}{2}=-6\ \to\ \boxed{\mathrm{a_1+a_4=-3}}\ \mathbf{(II)}}$

$\mathbf{Utilizando\ (II)\ em\ (I),\ obteremos:}\\\\ \mathrm{a_4=-3-a_1\ \to\ a_4=-(3+a_1)}\\\\ \mathrm{a_1.a_4=-54\ \to\ a_1.[-(3+a_1)]=-54\ \to}\\\\ \mathrm{\to\ a_1(3+a_1)=54\ \to\ a_1^2+3a_1-54=0}\\\\ \mathrm{a_1=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-3\pm\sqrt{3^2-4.1.(-54)}}{2.1}=}\\\\ \mathrm{=\dfrac{-3\pm\sqrt{9+216}}{2}=\dfrac{-3\pm\sqrt{225}}{2}=\dfrac{-3\pm15}{2}=\boxed{6}\ ou\ \boxed{-9}}$

$\mathbf{Para\ a_1=6:}\\\\ \mathrm{a_4=-(3+a_1)=-(3+6)=-9\ \to\ \boxed{\mathrm{a_4=-9}}}\\\\ \mathrm{a_4=a_1+3r\ \to\ -9=6+3r\ \to\ \boxed{\mathrm{r=-5}}}\\\\ \boxed{\mathrm{PA=(6,1,-4,-9)}}\\\\ \mathbf{Para\ a_1=-9:}\\\\ \mathrm{a_4=-(3+a_1)=-(3-9)=6\ \to\ \boxed{\mathrm{a_4=6}}}\\\\ \mathrm{a_4=a_1+3r\ \to\ 6=-9+3r\ \to\ \boxed{\mathrm{r=5}}}\\\\ \boxed{\mathrm{PA=(-9,-4,1,6)}}\\\\ \textbf{Solu\c{c}\~oes:}\\\\ \boxed{\mathrm{PA=\{(6,1,-4,-9);(-9,-4,1,6)\}}}$