Question

qual a soma dos cinco primeiros termos da P. G. (4, 12, ...)?

Answer 1

$P. G. (4, 12, ...)\\\\\\ S_n=?\\ a_1=4\\ n=5\\ q= \frac{12}{4} \to 3\\\\\\ S_n= \frac{a_1*(q^n-1)}{q-1} \\\\\\ S_5= \frac{4*(3^5-1)}{3-1} \to \\\\ S_5= \frac{4*(243-1)}{2} \to \\\\ S_5= \frac{4*242}{2} \to \\\\ S_5= \frac{968}{2} \to \\\\ S_n=484$



R.: A soma 5 primeiros termos dessa PG é igual a 484.

Espero ter ajudado...