Question

Quantos termos iniciais da P.G (-1,5,-25,...) devemos somar para obter a soma igual a 2604?

Answer 1

$q= \frac{a_n}{a_{n-1}} \\ \\ q= \frac{a_2}{a_{2-1}} \\ \\ q= \frac{a_2}{a_1} \\\\ q= \frac{5}{-1} \\ \\ q=-5 \\ \\ \\ S_n= \frac{a_1(q^n-1)}{q-1} \\ \\ 2604= \frac{-1((-5)^n-1)}{-5-1} \\ \\ 2604= \frac{-1(-5)^n+1}{-6}\\\\ 2604(-6)=-1(-5)^n+1\\\\-15624=-1(-5)^n+1\\\\1(-5)^n=15624+1\\\\(-5)^n=15625\\\\(-5)^n=(-5)^6\\\\n=6$

6 termos iniciais devem ser somados.


Renato.