Question

Determinar os termos da PG 2,6...486

Answer 1

$PG~(2, 6, ..., 486)\\\\a_1=2;\\ a_n=486; \\q= \frac{6}{2} =3\\n=?\\\\\\ a_n=a_1\times q^{n-1}\\\\\\ 486=2\times 3^{n-1}\to \\\\ \dfrac{486}{2} =3^{n-1}\to \\ \\243=3^{n-1}\to \\\\3^5=3^{n-1}\to \\\\5=n-1\to \\\\5+1=n\to\\\\ \boxed{n=6}\\\\\\\\ PG~(2, 6, 18, 54, 162, 486)$

Answer 2

a1 = 2
a2 = 6
q = a2/a1
q = 6/2
q = 3
an = 486
n = ?

               n - 1
an = a1.q

               n - 1
486 = 2.3

               n - 1
486/2 = 3

            n - 1
243 = 3

   5           n - 1
3     =   3

5 = n - 1
5 + 1 = n
6 = n
n = 6

Resp.: 6 termos.
--------------------------------------------------
243: 3
81: 3
27: 3
9: 3
3: 3
1      =  3.3.3.3.3 = 3⁵

A PG é: (2, 6, 18, 54, 162, 486)