Problem: 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 with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$ Solution: We can see $ \begin{align*} a_1+a_2+\ldots+a_{98} &= (a_1+a_3+\ldots+a_{97})+(a_2+a_4+\ldots+a_{98}) \\ & = [(a_2-1)+(a_4-1)+\ldots+(a_{98}-1)]+(a_2+a_4+\ldots+a_{98}) \\ & = 2(a_2+a_4+\ldots+a_{98})-49 \end{align*} $ Thus, the answer is $\frac{137+49}{2} = 93.$