Question

A soma de todos os termos de uma P.G finita e a 1020. Sendo a1=4 e q=2 calcule o numero de termos da P.G .

Answer 1

Sn = _a1 (q^n - 1)_     [fração]
                q - 1

Sn = 1020
a1 = 4 
q = 2
Substituindo os valores na equação:  

1020 = _4 (2^n -1)_
                2 - 1

1020 = 4 . (2^n - 1)           (:4)
255 = 2^n - 1
2^n = 256
2^n = 2^8
n = 8

Answer 2

$S_{n}= \dfrac{a_1.(q^n-1)}{q-1} \\\\\\S_n=1020\\a_1=4\\q=2\\\\1020= \dfrac{4.(2^n-1)}{2-1}\\\\ \dfrac{1020}{4}=2^n-1\\\\ 255 = 2^n - 1\\\\2^n=255+1\\\\2^n=256\\\\256|2\\128|2\\64\,\,\,|2\\32\,\,\,|2\\16\,\,\,|2\\8\,\,\,\,\,\,|2\\4\,\,\,\,\,\,|2\\2\,\,\,\,\,\,|2\\1\\\\256=2^8\\\\2^n=2^8\\\\\boxed{n=8}$