Question

Escreva a PG de três termos consecutivos, sabendo que a soma desses nº é
7 e o produto é – 27.

Answer 1

$PG~(a_{1},a_{2},a_{3})$

$a_{2}=a_{1}*q\\a_{3}=a_{1}*q^{2}$
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$a_{1}+a_{2}+a_{3}=7\\a_{1}+(a_{1}*q)+(a_{1}*q^{2})=7$

Colocando a₁ em evidência:

$a_{1}*(1+q+q^{2})=7$

O produto dos 3 é -27:

$a_{1}*a_{2}*a_{3}=-27\\a_{1}*a_{1}*q*a_{1}*q^{3}=-27\\(a_{1})^{3}*q^{3}=-27$

Raiz cúbica nos 2 lados da equação:

$\sqrt[3]{(a_{1})^{3}*q^{3}}=\sqrt[3]{-27}\\a_{1}*q=-3\\a_{1}=-3/q$
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$a_{1}*(1+q+q^{2})=7\\(-3/q)*(1+q+q^{2})=7\\(3/q)*(1+q+q^{2})=-7\\3*(1+q+q^{2})=-7q\\3+3q+3q^{2}=-7q\\3q^{2}+3q+7q+3=0\\3q^{2}+10q+3=0$

$\Delta=b^{2}-4ac\\\Delta=10^{2}-4*3*3\\\Delta=100-36\\\Delta=64\\\\q=\dfrac{-b\pm\sqrt{\Delta}}{2a}~~\therefore~~\dfrac{-10\pm\sqrt{64}}{2*3}~~\therefore~~\dfrac{-10\pm8}{6}~~\therefore~~\dfrac{-5\pm4}{3}$

$q'=\dfrac{-5+4}{3}=-\dfrac{1}{3}\\\\\\q''=\dfrac{-5-4}{3}=-3$
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Considerando q = -1/3:

$a_{1}=-3/q\\a_{1}=-3/(-1/3)\\a_{1}=3/(1/3)\\a_{1}=3*3/1\\a_{1}=9$

$a_{2}=a_{1}*q=9*(-1/3)=-3\\a_{3}=a_{2}*q=-3*(-1/3)=1$

$\boxed{P.G~(9,-3,~1)}$

Considerando q = -3:

$a_{1}=-3/q=-3/(-3)=1\\a_{2}=a_{1}*q=1*(-3)=-3\\a_{3}=a_{2}*q=(-3)*(-3)=9$

$\boxed{P.G~(1,-3,~9)}$