Question

Determine a soma dos termos da PG (3, -6, 12, ..., 768)

Answer 1

Explicação passo-a-passo:

Temos a seguinte PG(3, -6, 12, ..., 768)

Calculando a razão da PG:

$q=\dfrac{a_{2} }{a_{1} } \\\\q=\dfrac{-6}{3} \\\\\boxed{q=-2}$

Calculando a posição do último termo:

$a_{1}.q^{n-1}=a_{n} \\\\3.(-2)^{n-1}=768\\\\(-2)^{n-1}=\dfrac{768}{3} \\\\(-2)^{n-1}=256\\\\(-2)^{n-1}=(-2)^{8}\\\\\text Igualando\ os\ expoentes:\\\\n-1=8\\\\\boxed{n=9}$

Calculando a soma dos 9 termos da PG:

$S_{n}=\dfrac{a_{1}.(q^{n}-1) }{q-1} \\\\S_{9}=\dfrac{a_{1}.(q^{9}-1) }{q-1}\\\\S_{9}=\dfrac{3.((-2)^{9}-1) }{(-2)-1}\\\\S_{9}=\dfrac{3.(-512-1) }{-3}\\\\S_{9}=\dfrac{(-513) }{-1}\\\\\boxed{\boxed{S_{9}=513}}$