Question

A soma dos quadrados de três números em P.A decrescente é 126 e o 1º termoé igual a soma dos outros dois.Qual é essa P.A 

Answer 1

$\mathrm{PA=\{a_1,a_2,a_3\}=\{a_1,a_1+r,a_1+2r\}\ \to\ r\ \textless \ 0}\\ \mathrm{a_1^2+a_2^2+a_3^2=126\ \ \| \ a_1=a_2+a_3}\\\\ \mathrm{a_1=a_2+a_3\ \to\ a_1=a_1+r+a_1+2r}\\ \mathrm{a_1=2a_1+3r\ \to\ a_1-2a_1=3r\ \to\ a_1=-3r}\\\\ \mathrm{PA=\{-3r,-3r+r,-3r+2r\}=\{-3r,-2r,-r\}}\\\\ \mathrm{a_1^2+a_2^2+a_3^2=126\ \to\ (-3r)^2+(-2r)^2+(-r)^2=126}\\ \mathrm{9r^2+4r^2+r^2=126\ \to\ 14r^2=126\ \to\ r^2=\dfrac{126}{14}}\\ \mathrm{r^2=9\ \to\ r=\pm\sqrt{9}\ \to\ r=\pm3\ \to\ \mathbf{r=-3}}$

$\mathrm{PA=\{-3.(-3),-2.(-3),-1.(-3)\}=\mathbf{\{9,6,3\}}}$