The sequence given is,
[tex]8,4,2,\ldots[/tex]
Let us look for the common ration (r)
[tex]\begin{gathered} r=\frac{\text{second term}}{\text{first term}}=\frac{4}{8}=\frac{1}{2} \\ \therefore r=\frac{1}{2} \end{gathered}[/tex]
Given that
[tex]a=first\text{ term=8}[/tex]
The formula for the sum of sum of a finite geometric series is
[tex]S_n=\frac{a_1-a_{1r^n}}{1-r}[/tex]
We are to look for sum of first 6 terms
[tex]n=6[/tex]
Hence,
[tex]\begin{gathered} S_6=\frac{8-8\times(\frac{1}{2})^6}{1-\frac{1}{2}}=\frac{8-8\times\frac{1}{64}}{\frac{1}{2}}=\frac{8-\frac{1}{8}}{\frac{1}{2}} \\ S_6=\frac{7\frac{7}{8}}{\frac{1}{2}}=\frac{63}{8}\div\frac{1}{2}=\frac{63}{8}\times\frac{2}{1}=\frac{126}{8}=15.75 \end{gathered}[/tex]
Therefore, the sum of 6 first term is 15.75.
