Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$, $a_2$, $a_3\ldots$ is an [arithmetic progression](https://artofproblemsolving.com/wiki/index.php/Arithmetic_progression) with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$.
One approach to this problem is to apply the formula for the sum of an [arithmetic series](https://artofproblemsolving.com/wiki/index.php/Arithmetic_series) in order to find the value of $a_1$, then use that to calculate $a_2$ and sum another arithmetic series to get our answer.
A somewhat quicker method is to do the following: for each $n \geq 1$, we have $a_{2n - 1} = a_{2n} - 1$.  We can substitute this into our given equation to get $(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137$.  The left-hand side of this equation is simply $2(a_2 + a_4 + \ldots + a_{98}) - 49$, so our desired value is $\frac{137 + 49}{2} = \boxed{093}$.