Question

Dada a pg (1, √2/2, 1/2, ...) determine seu 11° termo.

Answer 1

$a_{n}=a_{1}.q^{n-1}$
$a_{n} = 1 . (\frac{1}{ \sqrt{2} })^{n-1}$
$a_{11} = 1 . (\frac{1}{ \sqrt{2} })^{10}$
$a_{11}=1. \frac{1}{32} = \frac{1}{32}$