Question

Qual é a ordem do termo igual a 125 na pg (1/25, 1/5, 1...)

Answer 1

Vamos começar determinando a razão:

$razao\,(q)~=~\frac{a_2}{a_1}\\\\\\razao\,(q)~=~\frac{\frac{1}{5}}{\frac{1}{25}}\\\\\\razao\,(q)~=~\frac{1~.~25}{5~.~1}\\\\\\\boxed{razao\,(q)~=~5}$

Agora, utilizando a equação do termo geral da PG:

$a_n~=~a_1~.~q^{n-1}\\\\\\125~=~\frac{1}{25}~.~5^{n-1}\\\\\\5^3~=~25^{-1}~.~5^{n-1}\\\\\\5^3~=~\left(5^2\right)^{-1}~.~5^{n-1}\\\\\\5^3~=~5^{-2}~.~5^{n-1}\\\\\\5^3~=~5^{n-1+(-2)}\\\\\\5^3~=~5^{n-3}\\\\\\3~=~n-3\\\\\\n~=~3+3\\\\\\\boxed{n~=~6}$

Answer 2

resolução!

q = a2 / a1

q = 1/5 ÷ 1/25

q = 1/5 * 25/1

q = 25/5

q = 5

an = a1 * q^n - 1

125 = 1/25 * 5^n - 1

125 ÷ 1/25 = 5^n - 1

125 * 25/1 = 5^n - 1

3125 = 5^n - 1

5^5 = 5^n - 1

n - 1 = 5

n = 5 + 1

n = 6