Question

numero de termos da pg ( 3,1/3,1/27,...,1/19683)

Answer 1

r = (1/3) / 3 = 1/9 
An = A1.q^(n-1)

1/19683 = 3.(1/9)^(n-1)
1/59049 = (1/9)^(n-1)
Log 1/59049 = log (1/9)^(n-1)
log 1 - log 59049 = (n-1).(log 1 - log 9)
0 - 4,77 = (n-1)(0 - 0,9542)
-4,77 = -0,9542n + 0,9542
0,9542n = 0,9542 + 4,77
0,9542n = 5,7242
n = 6

Espero ter ajudado