Answer
True statement
Work Step by Step
$10-5+\frac{5}{2}-\ldots $
The limit of the infinite geometric series is given by
${{s}_{\infty }}=\frac{{{a}_{1}}}{1-r}$
When $\left| r \right|<1$ then the limit exists and if $\left| r \right|\ge 1$, no limit exists.
To check the limit find the common ratio
$r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$
Put $n=1$ in $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$,
$\begin{align}
& r=\frac{{{a}_{1+1}}}{{{a}_{1}}} \\
& =\frac{{{a}_{2}}}{{{a}_{1}}}
\end{align}$
Put ${{a}_{1}}=10$ and ${{a}_{2}}=-5$ in $r=\frac{{{a}_{2}}}{{{a}_{1}}}$
$\begin{align}
& r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
& =\frac{-5}{10} \\
& =-\frac{1}{2}
\end{align}$
And,
$\begin{align}
& \left| r \right|=\left| -\frac{1}{2} \right| \\
& =\frac{1}{2}
\end{align}$
Thus, $\left| r \right|=\frac{1}{2}<1$, then the limit exists.
Therefore, the given statement is true.
