the third term of a geometric sequence is 6 and the sixth term is 3 over 32. Determine the first terms and the general term Tn?

Answer 1

1 a
2 ar
3 ar^2
4 ar^3
5 ar^4
6 ar^5

n ar^(n-1)

so
ar^2 = 6
ar^5 = 3/32
so
a = 6/r^2
(6/r^2)r^5 = 3/32
6/r^3 = 3/32
r^3 = 32*2 = 64
r = 4
then
a = 6/16 = 3/4
so
Tn = (3/4)4^(n-1)