If the sum of the first $m$ terms of an arithmetic series is $n$ and the sum of the first $n$ terms is $m$, the sum of the first $(m + n)$ terms is —

For an arithmetic series, the condition $S_m=n$ and $S_n=m$ is a standard result that forces $S_{m+n}=-(m+n)$. This follows because $S_m-S_n=n-m$ leads to $a+\frac{(m+n-1)}{2}d=-1$, which when used in $S_{m+n}=\frac{(m+n)}{2}[2a+(m+n-1)d]$ yields $-(m+n)$.