Answer
False
Work Step by Step
Let $a_n$ be the initial arithmetic series, $d_1$ its common difference, $b_n$ the second arithmetic series, $d_2$ its common difference and $n$ the number of terms in the sum.
We are given:
$$\begin{align*}
d_2&=2d_1\\
b_1&=a_1.
\end{align*}$$
The sum of each series is:
$$\begin{align*}
S_{n_1}&=\dfrac{n}{2}(a_1+a_n)\\
&=\dfrac{n}{2}(a_1+a_1+(n-1)d_1)\\
&=\dfrac{n}{2}(2a_1+(n-1)d_1).
\end{align*}$$
$$\begin{align*}
S_{n_2}&=\dfrac{n}{2}(b_1+b_n)\\
&=\dfrac{n}{2}(a_1+a_1+(n-1)d_2)\\
&=\dfrac{n}{2}(2a_1+2(n-1)d_1).
\end{align*}$$
Calculate $S_2-2S_1$:
$$\begin{align*}
S_2-2S_1&=\dfrac{n}{2}(2a_1+2(n-1)d_1)-2\cdot \dfrac{n}{2}(2a_1+(n-1)d_1)\\
&=-na_1\not=0.
\end{align*}$$
Since $S_1-2S_2\not=0$ it means that the sum of the series is not doubled.
The given statement is FALSE.
