Answer
$${a_4} = 1,\,\,\,\,\,{a_n} = 27{\left( {\frac{1}{3}} \right)^{n - 1}}{\text{ and }}{S_5} = \frac{{121}}{3}$$
Work Step by Step
$$\eqalign{
& {a_1} = 27,\,\,\,\,r = 1/3 \cr
& {\text{The general term of a geometric sequence is }}{a_n} = {a_1}{r^{n - 1}}.{\text{ }} \cr
& {\text{substituting }}{a_1} = 27,\,\,\,\,r = 1/3 \cr
& {a_n} = 27{\left( {\frac{1}{3}} \right)^{n - 1}} \cr
& \cr
& {\text{find }}{a_4},{\text{ substitute }}n = 4{\text{ into the general term formula}} \cr
& {a_4} = 27{\left( {\frac{1}{3}} \right)^{4 - 1}} \cr
& {a_4} = 1 \cr
& \cr
& {\text{then the sum of the first }}n{\text{ terms}}{\text{, is given by}} \cr
& {S_n} = \frac{{{a_1}\left( {{r^n} - 1} \right)}}{{r - 1}},{\text{ where }}r \ne 1 \cr
& {\text{For the first five terms we have}} \cr
& {S_5} = \frac{{\left( {27} \right)\left( {{{\left( {1/3} \right)}^5} - 1} \right)}}{{1/3 - 1}} \cr
& {\text{by a calculator}} \cr
& {S_5} = \frac{{121}}{3} \cr
& \cr
& {\text{then}} \cr
& {a_4} = 1,\,\,\,\,\,{a_n} = 27{\left( {\frac{1}{3}} \right)^{n - 1}}{\text{ and }}{S_5} = \frac{{121}}{3} \cr} $$
