Answer
$${S_5} = \frac{{33}}{4}$$
Work Step by Step
$$\eqalign{
& {a_1} = 12,{a_2} = - 6,{a_3} = 3,{a_4} = - 3/2,... \cr
& {\text{to verify that the sequence is geometric}}{\text{, divide each term except the first by }} \cr
& {\text{the preceding term}} \cr
& \frac{{{a_4}}}{{{a_3}}} = \frac{{ - 3/2}}{3} = - \frac{1}{2} \cr
& \frac{{{a_3}}}{{{a_2}}} = \frac{3}{{ - 6}} = - \frac{1}{2} \cr
& \frac{{{a_2}}}{{{a_1}}} = \frac{{ - 6}}{{12}} = - \frac{1}{2} \cr
& {\text{the ratio is constant}}{\text{, so the sequence is geometric with }}r = - \frac{1}{2} \cr
& {\text{then the sum of the first }}n{\text{ terms}}{\text{ is given by}} \cr
& {S_n} = \frac{{{a_1}\left( {{r^n} - 1} \right)}}{{r - 1}},{\text{ where }}r \ne 1 \cr
& {\text{let }}n = 5,\,\,\,{a_1} = 12{\text{ and }}r = - \frac{1}{2} \cr
& {S_5} = \frac{{\left( {12} \right)\left( {{{\left( { - \frac{1}{2}} \right)}^5} - 1} \right)}}{{ - \frac{1}{2} - 1}} \cr
& {\text{simplifying}} \cr
& {S_5} = \frac{{\left( {12} \right)\left( { - \frac{{33}}{{32}}} \right)}}{{ - \frac{3}{2}}} \cr
& {S_5} = \frac{{33}}{4} \cr} $$
