Answer
$1140$.
Work Step by Step
When we consider the sequences provided,
$\begin{align}
& \{{{a}_{n}}\}=-5,10,-20,40,\ldots \\
& \{{{b}_{n}}\}=10,-5,-20,-35,\ldots \\
& \{{{c}_{n}}\}=-2,1,-\frac{1}{2},\frac{1}{4},\ldots
\end{align}$
We get ${{a}_{1}}=-5,{{a}_{2}}=10,{{a}_{3}}=-20$.
So we can clearly observe that,
$\begin{align}
& \frac{{{a}_{2}}}{{{a}_{1}}}=\frac{{{a}_{3}}}{{{a}_{2}}} \\
& =-2
\end{align}$
Thus, this is a geometric sequence with the common ratio being $ r=-2$.
The sum of the first $ n $ terms of a geometric sequence is ${{S}_{n}}=\frac{a\times \left( 1-{{r}^{n}} \right)}{1-r}$ where $ a={{a}_{1}}$.
Now;
$\begin{align}
& {{S}_{9}}=\frac{-5\times \left( 1-{{\left( -2 \right)}^{9}} \right)}{1-\left( -2 \right)} \\
& =\frac{-5\times \left( 1+512 \right)}{1+2} \\
& =\frac{-5\times \left( 513 \right)}{3} \\
& =-855
\end{align}$
Likewise, ${{c}_{1}}=-2,{{c}_{2}}=1,{{c}_{3}}=-\frac{1}{2}$.
Now we can see that,
$\begin{align}
& r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
& =\frac{1}{-2} \\
& =-\frac{1}{2}
\end{align}$
Also,
$\begin{align}
& r=\frac{{{a}_{3}}}{{{a}_{2}}} \\
& =\frac{-\frac{1}{2}}{1} \\
& =-\frac{1}{2}
\end{align}$
Thus, this is a geometric sequence with common ratio being $ r=-\frac{1}{2}$.
The general term of an arithmetic sequence is $ S'{{'}_{\infty }}=\frac{c}{1-r}$ where $ c={{c}_{1}}$.
So;
$\begin{align}
& S'{{'}_{\infty }}=\frac{-2}{1-\left( -\frac{1}{2} \right)} \\
& =\frac{-2}{1+\frac{1}{2}} \\
& =\frac{-2}{\frac{2+1}{2}} \\
& =-\frac{4}{3}
\end{align}$
Hence, simply multiply both equations,
$\begin{align}
& {{S}_{9}}\times S'{{'}_{\infty }}=-855\times \left( -\frac{4}{3} \right) \\
& =\frac{855\times 4}{3} \\
& =\frac{3420}{3} \\
& =1140
\end{align}$
