Question

Qual é o termo geral de cada sequencia dada:
A) ( 3,6,12,24,48,...)
B) ( 2,6,18,54,162,...)

Me ajudeeem é pra amanha

Answer 1

$Todas~~duas~~s\~ao~~PG\\\\\\ a_n=a_1\times q^{n-1}\\\\\\ A) ~~( 3,6,12,24,48,...)\\\\q\to~6:3=2\\\\\\ a_n= 3\times 2^{n-1}\to~~~~ \large\boxed{ \dfrac{a_n}{3}= 2^{n-1}}\\\\\\\\\\\\ B) ~~( 2,6,18,54,162,...) \\\\q\to~6:2=3\\\\\\ a_n=2\times 3^{n-1}\to~~~~ \large\boxed{ \dfrac{a_n}{2}= 3^{n-1}}$

Answer 2

PG: Progressão Geométrica
              n - 1
an =a1.q

               2 - 1
a2 = a1.q

a2 = a1.q
a2 = q
a1

q = a2
      a1
--------------------------------
a)
A) ( 3,6,12,24,48,...)

a1 = 3
a2 = 6

q = a2/a1 
q = 6/3
q = 2

               n - 1
an = a1.q

Resp.: 
Termo Geral:

             n - 1
an = 3.2
=======================
b)

B) ( 2,6,18,54,162,...)

a1 = 2
a2 = 6

q = a2/a1
q = 6/2
q = 3
                n - 1
an = a1.q

Resp.: 
Termo Geral:

             n - 1
an = 2.3