For geometric progression you'll have the formula:
[tex]a_n = a_1 r ^{n-1} [/tex]
where:
a1 is the first term = 3
an is the last term which in this case is a6=32/81
r is the common ratio
substituting the formula:
[tex]a_n = a_1 r^{n-1} [/tex]
[tex] \frac{32}{81} = 3 r ^{6-1} [/tex]
[tex] \frac{32}{81(3)} =r^5[/tex]
extracting the fifth root you'll have
[tex]r = \frac{2}{3} [/tex]
[tex]a_2 = 3( \frac{2}{3} )[/tex]
[tex]a_2 = 2[/tex]
[tex]a_3 = 2( \frac{2}{3}) [/tex]
[tex]a_3 = \frac{4}{3} [/tex]
[tex]a_4 = \frac{4}{3} (\frac{2}{3})[/tex]
[tex]a_4 = \frac{8}{9}[/tex]
[tex]a_5 = \frac{8}{9} ( \frac{2}{3} )[/tex]
[tex]a_5 = \frac{16}{27} [/tex]
therefore you'll have the geometric series as:
3, 2, 4/3, 8/9, 16/27, 32/81
