Question

4) Encontre o termo geral da P.A. (12, 16, 20,…).

Answer 1

Olá,

$a_{1} = 12$

$r = 16 - 12 = 4$

Assim:

$a_{n} = a_{1} + (n - 1)r$

$a_{n} = 12 + (n - 1)(4)$

$a_{n} = 12 + 4n - 4$

$a_{n} = 4n + 8$