The number $2.5252525\ldots$ can be written as a fraction. 
When reduced to lowest terms the sum of the numerator and denominator of this fraction is:
$\textbf{(A) }7\qquad \textbf{(B) }29\qquad \textbf{(C) }141\qquad \textbf{(D) }349\qquad \textbf{(E) }\text{none of these}$
Solution 1
Let $x = 2.5252525\ldots$ , so
\[100x = 252.52525\ldots\]
\[x = 2.5252525\ldots\]
\[99x = 250\]
\[x = \frac{250}{99}\]
The sum of the numerator and the denominator is $349$, so the answer is $\boxed{\textbf{(D)}}$ .

Solution 2
The number $2.5252525\ldots$ can also be written as
\[2+\frac{5}{10} + \frac{2}{100} + \frac{5}{1000} + \frac{2}{10,000} \ldots\]
This is really two infinite series -- one with first term $2$ and common difference $\frac{1}{100}$ and one with first term $\frac{5}{10}$ and common difference $\frac{1}{100}$.  Apply the infinite series formula $S = \frac{a_1}{1-r}$ on both series.
\[\frac{2}{1-\frac{1}{100}} + \frac{\frac{5}{10}}{1-\frac{1}{100}}\]
\[\frac{2}{\frac{99}{100}} + \frac{\frac{5}{10}}{\frac{99}{100}}\]
\[2 \cdot \frac{100}{99} + \frac{5}{10} \cdot \frac{100}{99}\]
\[\frac{200}{99} + \frac{50}{99}\]
\[\frac{250}{99}\]
The sum of the numerator and the denominator is $349$, so the answer is $\boxed{\textbf{(D)}}$ .