Na PG (y,2y+2,3y+3) , o quarto termo , que é diferente de zero.... Pergunta de ideia dejaruaba20

October 2020 1 136 Report

Na PG (y,2y+2,3y+3) , o quarto termo , que é diferente de zero

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Temos que:

$\boxed{a_2^2=a_1*a_3}\\\\ (2y+2)^2 = y(3y+3)\\\\ 4y^2+8y+4 = 3y^2+3y\\\\ y^2+5y+4=0\\\\ \Delta = 25-16\\\\ \Delta = 9$

$x = \frac{-5\ +\ ou\ -\ 3}{2}\\\\ x' = \frac{-5+3}{2} =\frac{-2}{2} = -1\\\\ x''=\frac{-5-3}{2}=\frac{-8}{2} = -4\\\\ \boxed{S(-4,-1)}$

Agora, podemos substituir qualquer desse valores de y para encontrar os termos da PG

$a_1 = y = -4\\\\ a_2 = 2y+2 = -6\\\\ a_3 = 3y+3=-9$

A razão dessa PG é:

$q = \frac{a_2}{a_1}\\\\ q = \frac{-6}{-4}\\\\ \boxed{q=\frac{3}{2}}$

Para descobrir o quarto termo, multiplicamos o terceiro termo pela razão.

$a_4 = a_3*q\\\\ a_4 = -9*\frac{3}{2}\\\\ \boxed{a_4=-\frac{27}{2}}$

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