Question

Determine a soma dos 8 primeiros termos de cada PG.
a) (3,12,48...)
b) (-4,8,-16,...)

Answer 1

$S_n=\dfrac{a_1\cdot(q^{n}-1)}{q-1}$

a) $(3,12,48,\dots)$

$a_1=3$ e $q=\dfrac{12}{3}=4$

$S_8=\dfrac{3\cdot(4^8-1)}{4-1}=\dfrac{3\cdot(65~536-1)}{3}=65~535$.

b) $(-4,8,-16,\dots)$

$a_1=-4$ e $q=\dfrac{8}{-2}=-2$

$S_8=\dfrac{(-4)\cdot[(-2)^8-1]}{-2-1}=\dfrac{(-4)\cdot[256-1]}{-3}=\dfrac{(-4)\cdot255}{-3}$

$S_8=\dfrac{-1~020}{-3}=340$