Question

determine três numeros em PG de modo que sua soma seja 7 e seu produto seja 64/27

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\begin{cases}\mathsf{a_1 + a_2 + a_3 = 7}\\\\\mathsf{a_1.a_2.a_3 = \dfrac{64}{27}}\\\\\mathsf{\dfrac{a_3}{a_2} = \dfrac{a_2}{a_1}}\end{cases}$

$\mathsf{a_1.a_3 = (a_2)^2}$

$\mathsf{a_1.a_3.a_2 = \dfrac{64}{27}}$

$\mathsf{(a_2)^3 = \dfrac{64}{27}}$

$\mathsf{a_2 = \dfrac{4}{3}}$

$\mathsf{a_1 + a_3 = 7 - \dfrac{4}{3}}$

$\mathsf{a_1 + a_3 = \dfrac{21 - 4}{3}}$

$\mathsf{a_1 + a_3 = \dfrac{17}{3}}$

$\begin{cases}\mathsf{a_1 + a_3 = \dfrac{17}{3}}\\\\\mathsf{a_1.a_3 = \dfrac{16}{9}}\end{cases}$

$\mathsf{a_1 = \dfrac{17}{3} - a_3}$

$\mathsf{a_1 = \dfrac{17 - 3a_3}{3}}$

$\mathsf{\left(\dfrac{17 - 3a_3}{3}\right).a_3 = \dfrac{16}{9}}$

$\mathsf{\left(\dfrac{17 - 3a_3}{1}\right).a_3 = \dfrac{16}{3}}$

$\mathsf{51.a_3 - 9.(a_3)^2 = 16}$

$\mathsf{9.(a_3)^2 - 51.a_3 + 16 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (-51)^2 - 4.9.16}$

$\mathsf{\Delta = 2.601 - 576}$

$\mathsf{\Delta = 2.025}$

$\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{51 \pm \sqrt{2.025}}{18} \rightarrow \begin{cases}\mathsf{x' = \dfrac{51 + 45}{18} = \dfrac{96}{18} = \dfrac{16}{3}}\\\\\mathsf{x'' = \dfrac{51 - 45}{18} = \dfrac{6}{18} = \dfrac{1}{3}}\end{cases}}$

$\mathsf{a_1 = \dfrac{17}{3} - \dfrac{1}{3}}$

$\mathsf{a_1 = \dfrac{16}{3}}$

$\boxed{\boxed{\mathsf{S = \left\{a_1 = \dfrac{16}{3};\:a_2 = \dfrac{4}{3};\:a_3 = \dfrac{1}{3}\right\}}}}$