Question

Calcule a soma da PG (6, 3, ...)

Answer 1

q=3/6=1/2
S = a1/(1-q) quando | q | < 1
S=6/1-1/2=6/1/2=12

Answer 2

$a_{1}=6$
$a_{2}=3$

$q=a_{2}/a_{1}=3/6=1/2$

$0 < |q| < 1:$ Soma infinita

$S_{n} = a_{1}/(1 - q)$
$S_{n} = 6/(1 - [1/2])$
$S_{n} = 6/([2/2] - [1/2])$
$S_{n} = 6/([2-1]/2)$
$S_{n} = 6/(1/2)$
$S_{n} = 6*2/1$
$S_{n}=12$