Question

quantos termos devemos considerar na PG 3,6,...para obter uma soma de 765

Answer 1

6/3= 2, a razao é 2!

Sn=a1{(q^n)-1}/q-1
765=3*{(2^n)-1}/1

posso dividir as parcelas por 3:
255=(2^n)-1
256=2^n

decomponho o 256:
2^8=2^n

"corto as bases"
n=8

Answer 2

$q=\dfrac{a_2}{a_1}\\\\\\q=\dfrac{6}{2}\\\\\\q=2\\\\\\\\S_n=\dfrac{a_1(q^n-1)}{q-1}\\\\\\756=\dfrac{3(2^n-1)}{2-1}\\\\\\756=\dfrac{3(2^n-1)}{1}\\\\\\756=3(2^n-1)\\\\\\\dfrac{756}{3}=2^n-1\\\\\\255=2^n-1\\\\255+1=2^n\\\\2^n=256\\\\2^n=2^8\\\\\therefore\\\\\boxed{n=8}\\\\\text{Devemos considerar 8 termos.}$