Answer
10.5
Work Step by Step
The series is $16-8+4-\cdots $. Here, ${{a}_{1}}=16$, $n=6$.
The value of r is calculated below:
$\begin{align}
& r=\frac{-8}{16} \\
& =-\frac{1}{2}
\end{align}$
Substituting ${{a}_{1}}=6$, $n=9$ and $r=2$ in ${{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r}$ , for any $r\ne 1$, we get
$\begin{align}
& {{S}_{9}}=\frac{16\left( 1-{{\left( -\frac{1}{2} \right)}^{6}} \right)}{1-\left( -\frac{1}{2} \right)} \\
& =\frac{16\left( 1-\frac{1}{64} \right)}{\frac{3}{2}} \\
& =\frac{16\cdot \left( -\frac{63}{64} \right)}{\frac{3}{2}} \\
& =10.5
\end{align}$
Thus, the sum of the first nine terms, ${{S}_{6}}$ for $16-8+4-\cdots $ is 10.5.
