When the sum of the first ten terms of an arithmetic progression is four times the sum of the first five terms, the ratio of the first term to the common difference is: 
$\textbf{(A)}\ 1: 2 \qquad \textbf{(B)}\ 2: 1 \qquad \textbf{(C)}\ 1: 4 \qquad \textbf{(D)}\ 4: 1 \qquad \textbf{(E)}\ 1: 1$
Let our first term be $a$ and our common difference be $d$. Thus, the first few terms of the sequence are $a$, $a + d$, $a + 2d$, ...
The sum of the first 5 terms is
\[a + (a + d) + (a + 2d) + ...  + (a + 4d) = 5a + 10d\]
The sum of the first 10 terms is 
\[a + (a + d) + (a + 2d) + ... + (a + 9d) = 10a + 45d\]
We are told that the latter is four times the former. Hence, 
\[4(5a + 10d) = 10a + 45d\]
\[20a + 40d = 10a + 45d\]
\[10a = 5d\]
\[\frac{a}{d}  = \frac{1}{2}\]
Therefore, our answer is $\fbox{(A) 1:2}$