jmc

Let $S=a_1+a_3+\cdots +a_{97}$ and $T=a_2+a_4+\cdots +a_{98}$. Then the given equation states that $S+T=137$, and we want to find $T$.

We can build another equation relating $S$ and $T$: note that $$\begin{array}{rl}T-S& =(a_2-a_1)+(a_4-a_3)+\cdots +(a_{98}-a_{97})\\ & =\underset{49\text{ times }}{\underbrace{1+1+\cdots +1}}\\ & =49\end{array}$$since $(a_n)$ has common difference $1$. Then, adding the two equations $S+T=137$ and $T-S=49$, we get $2T=137+49=186$, so $T=\frac{186}{2}=\boxed{93}$.