Question

determine a soma da PG 1/3,2/9,4/27

Answer 1

$P.G(1/3,2/9,4/27,...)$

$a_{1} = 1/3$
$a_{2}=2/9$

$q = a_{2}/a_{1} => (2/9)/(1/3) => (2/9)*(3/1) => (2/3)*(1/1) => q = 2/3$

$0 < q < 1$: Soma infinita

$S_{n} = a_{1} / (1 - q)$
$S_{n} = (1/3) / (1 - [2/3])$
$S_{n} = (1/3)/ ([3/3]-[2/3])$
$S_{n} = (1/3)/([3-2]/3)$
$S_{n} = (1/3)/(1/3)$
$S_{n}=1$

Answer 2

a1 =1/3 = 3^-1

q =  a2  =   2/9  ==> 2  . 3   ==> 2 
       a1       1/3         9     1          3

n= ?
an =?

Para determinarmos a soma precisamos saber qual o termo a ser calculado ou dando o último termos.