Question Find the sum of the first 6 terms of the following sequence. Round to the nearest hundredth if necessary. \[ 8, \quad 4, \quad 2, \ldots \] Sum of a finite geometric series: \[ S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r} \]

Let the first term be \( a_{1}=8 \) and the common ratio be \[ r=\frac{4}{8}=\frac{1}{2}. \] The sum of the first 6 terms is given by: \[ S_{6}=\frac{a_{1}-a_{1}r^{6}}{1-r}. \] Substitute \( a_{1}=8 \) and \( r=\frac{1}{2} \): \[ S_{6}=\frac{8-8\left(\frac{1}{2}\right)^{6}}{1-\frac{1}{2}}. \] Calculate \(\left(\frac{1}{2}\right)^{6}\): \[ \left(\frac{1}{2}\right)^{6}=\frac{1}{64}. \] So, \[ S_{6}=\frac{8-8\left(\frac{1}{64}\right)}{\frac{1}{2}}=\frac{8-\frac{8}{64}}{\frac{1}{2}}=\frac{8-\frac{1}{8}}{\frac{1}{2}}. \] Convert \(8\) to a fraction with denominator \(8\): \[ 8=\frac{64}{8}. \] Then, \[ \frac{64}{8}-\frac{1}{8}=\frac{63}{8}. \] Now, divide by \(\frac{1}{2}\): \[ S_{6}=\frac{\frac{63}{8}}{\frac{1}{2}}=\frac{63}{8}\times2=\frac{63}{4}. \] Finally, compute the fraction: \[ \frac{63}{4}=15.75. \] The sum of the first 6 terms is \(15.75\).