Posted by bahle on Thursday, January 23, 2014 at 2:58am.
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?

maths  Damon, Thursday, January 23, 2014 at 3:10am
1 a
2 ar
3 ar^2
4 ar^3
5 ar^4
6 ar^5
n ar^(n1)
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^(n1)