Question

3. Dados o primeiro termo - 5 e a razão 1/3, escreva a P.G. associada de 7 termos. *


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Answer 1

Resposta:

$\boxed{PG:\ (-5; \dfrac{-5}{\ \ 3};\dfrac{-5}{\ \ 9};\dfrac{-5}{\ 27};\dfrac{-5}{\ 81};\dfrac{-5}{243};\dfrac{-5}{729})}$

Explicação passo-a-passo:

Vamos extrair or dados:

A1 = -5

q = 1/3

n = 7

PG = ?

Fórmula:

$A_n=A_1\ .\ q^{(n\ -\ 1)}\\\\A_2=A_1\ .\ q^{(2\ -\ 1)}\\A_2=-5\ .\ {(\dfrac{1}{3})}^1 = \dfrac{-5}{\ \ 3}\\\\A_3=A_1\ .\ q^{(3\ -\ 1)}\\A_3=-5\ .\ {(\dfrac{1}{3})}^2 = \dfrac{-5}{\ \ 9}\\\\A_4=A_1\ .\ q^{(4\ -\ 1)}\\A_4=-5\ .\ {(\dfrac{1}{3})}^3 = \dfrac{-5}{\ 27}\\\\A_5=A_1\ .\ q^{(5\ -\ 1)}\\A_5=-5\ .\ {(\dfrac{1}{3})}^4 = \dfrac{-5}{\ 81}\\\\A_6=A_1\ .\ q^{(6\ -\ 1)}\\A_6=-5\ .\ {(\dfrac{1}{3})}^5 = \dfrac{-5}{243}\\\\A_7=A_1\ .\ q^{(7\ -\ 1)}\\A_7=-5\ .\ {(\dfrac{1}{3})}^6 = \dfrac{-5}{729}$

$A_1=-5\\A_2= \dfrac{-5}{\ \ 3}\\A_3=\dfrac{-5}{\ \ 9}\\A_4=\dfrac{-5}{\ 27}\\A_5=\dfrac{-5}{\ 81}\\A_6=\dfrac{-5}{243}\\A_7=\dfrac{-5}{729}\\\\\boxed{PG:\ (-5; \dfrac{-5}{\ \ 3};\dfrac{-5}{\ \ 9};\dfrac{-5}{\ 27};\dfrac{-5}{\ 81};\dfrac{-5}{243};\dfrac{-5}{729})}$

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