## Precalculus (6th Edition) Blitzer

$1140$.
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}