8.6. Let the function $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=a x+b, a, b \in \mathbb{R}^{*}$. If $m, n$ are distinct natural numbers such that

$$
\begin{gathered}
n>2, m>2, f(1)+f(2)+\cdots+f(m)=-n \text { and } f(1)+f(2)+\cdots+f(n)=-m \text {, calculate } \\
\frac{f(1)+f(2)+\cdots+f(m+n)}{m+n}
\end{gathered}
$$
Solution: $f(1)+f(2)+\cdots+f(m)=a \cdot \frac{m(m+1)}{2}+m b$, similarly $f(1)+f(2)+\cdots+f(n)=$ $a \cdot \frac{n(n+1)}{2}+n b \Rightarrow a m^{2}+a m+2 m b+2 n=0$ and $a n^{2}+a n+2 n b+2 m=0 \Rightarrow$ $a m^{2}+a m+2 m b+2 n-a n^{2}-a n-2 n b-2 m=0 \Leftrightarrow a(m+n+1)=2(1-b)$ $(m \neq n)$.

$f(1)+f(2)+\cdots+f(m+n)=a \cdot \frac{(m+n)(m+n+1)}{2}+b(m+n)$. However, $a(m+n+1)=$

$2(1-b) \Rightarrow f(1)+f(2)+\cdots+f(m+n)=\frac{(m+n) 2(1-b)}{2}+b(m+n)=m+n \Rightarrow$ $\frac{f(1)+f(2)+\cdots+f(m+n)}{m+n}=1$.

Answer: 1