Question

Dada a PG (1/3, 2/9, 4/27, ...) identifique a posição do termo 128/6.561

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{a_n = a_1.q^{n - 1}}$

$\mathsf{\dfrac{128}{6.561} = \left(\dfrac{1}{3}\right).\left(\dfrac{2}{3}\right)^{n - 1}}$

$\mathsf{\left(\dfrac{2}{3}\right)^{n - 1} = \dfrac{128}{2.187} }$

$\mathsf{\left(\dfrac{2}{3}\right)^{n - 1} = \left(\dfrac{2}{3}\right)^7}$

$\mathsf{n - 1 = 7}$

$\mathsf{n = 7 + 1}$

$\boxed{\boxed{\mathsf{n = 8}}}$