Question

Qual é o número de termos da PG(4,12,36,2916

Answer 1

$(a_1,a_2,a_3,\dots,a_n)=(4,12,36,\dots,2916)\\\\\text{Termo geral}\ \to\ \boxed{a_n=a_kq^{n-k}}\\\\ a_2=a_1q^{2-1}\ \therefore\ q=\dfrac{a_2}{a_1}=\dfrac{12}{4}\ \therefore\ q=3\\\\ a_n=a_1q^{n-1}\ \therefore\ a_n=4(3^{n-1})\\\\ \text{Para}\ a_n=2916\ \to\ 2916=4(3^{n-1})\ \therefore\\\\ 729=3^{n-1}\ \therefore\ 3^6=3^{n-1}\ \therefore\ 6=n-1\ \therefore\\\\ \boxed{n=7\ \text{termos}}$