Question

Quantos termos tem a PG (3125, 625, 125, ..., 1/625)?

Answer 1

Explicação passo-a-passo:

A razão dessa PG é:

$\sf q=\dfrac{a_2}{a_1}$

$\sf q=\dfrac{625}{3125}$

$\sf q=\dfrac{1}{5}$

Utilizando a fórmula do termo geral:

$\sf a_n=a_1\cdot q^{n-1}$

$\sf \dfrac{1}{625}=3125\cdot\left(\dfrac{1}{5}\right)^{n-1}$

$\sf \left(\dfrac{1}{5}\right)^{n-1}=\dfrac{\frac{1}{625}}{3125}$

$\sf \left(\dfrac{1}{5}\right)^{n-1}=\dfrac{1}{625}\cdot\dfrac{1}{3125}$

$\sf \left(\dfrac{1}{5}\right)^{n-1}=\left(\dfrac{1}{5}\right)^{4}\cdot\left(\dfrac{1}{5}\right)^{5}$

$\sf \left(\dfrac{1}{5}\right)^{n-1}=\left(\dfrac{1}{5}\right)^{4+5}$

$\sf \left(\dfrac{1}{5}\right)^{n-1}=\left(\dfrac{1}{5}\right)^{9}$

Igualando os expoentes:

$\sf n-1=9$

$\sf n=9+1$

$\sf n=10$