An [infinite](https://artofproblemsolving.com/wiki/index.php/Infinite) [geometric series](https://artofproblemsolving.com/wiki/index.php/Geometric_series) has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common [ratio](https://artofproblemsolving.com/wiki/index.php/Ratio) of the original series is $\frac mn$ where $m$ and $n$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime) [integers](https://artofproblemsolving.com/wiki/index.php/Integer). Find $m+n.$
Let's call the first term of the original [geometric series](https://artofproblemsolving.com/wiki/index.php/Geometric_series) $a$ and the common ratio $r$, so $2005 = a + ar + ar^2 + \ldots$.  Using the sum formula for [infinite](https://artofproblemsolving.com/wiki/index.php/Infinite) geometric series, we have $\;\;\frac a{1 -r} = 2005$.  Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \ldots$.  We know this series has sum  $20050 = \frac{a^2}{1 - r^2}$.  Dividing this equation by $\frac{a}{1-r}$, we get $10 = \frac a{1 + r}$.  Then $a = 2005 - 2005r$ and $a = 10 + 10r$ so $2005 - 2005r = 10 + 10r$, $1995 = 2015r$ and finally $r = \frac{1995}{2015} = \frac{399}{403}$, so the answer is $399 + 403 = \boxed{802}$.  
(We know this last fraction is fully reduced by the [Euclidean algorithm](https://artofproblemsolving.com/wiki/index.php/Euclidean_algorithm) -- because $4 = 403 - 399$, $\gcd(403, 399) | 4$.  But 403 is [ odd](https://artofproblemsolving.com/wiki/index.php/Odd_integer), so $\gcd(403, 399) = 1$.)