Question

Calcule a soma dos 8 primeiros termos da PG (3,6,9)

Answer 1

$P.G~(3,6,12,...)$

$a_{1}=3\\a_{2}=6\\\\q=a_{2}/a_{1}=6/3=2$
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$S_{n}=\dfrac{a_{1}(q^{n}-1)}{q-1}\\\\\\S_{8}=\dfrac{a_{1}(q^{8}-1)}{q-1}$

$S_{8}=\dfrac{3(2^{8}-1)}{2-1}\\\\\\S_{8}=3(256-1)\\\\S_{8}=3(255)\\\\\boxed{\boxed{S_{8}=765}}$