Show that there exists $c\in[a,b]$ such that $f(c)=\frac{1}{n} ( f(x_1)+f(x_2)+...f(x_n))$ 
Given $f$ a continuous function on $(a,b)$ such that $x_1,x_2,...,x_n$ $n$ elements of $(a,b)$, show that it exists $c\in[a,b]$ such that $f(c)=\frac{1}{n} ( f(x_1)+f(x_2)+...f(x_n))$

That equals to show that : $f(c)+f(c)+...+f(c) ( \text{n times})=f(x_1)+f(x_2)+...+f(x_n)$
There we assume the function $h(x)=f(x)-f(x_1)+f(x)-f(x_2)+...+f(x)-f(x_n)$
Its there in my mind to use the IVT but I don't know how in such serie function like that

Find $l,h \in \{1,2,\ldots ,n \}$ such that $f(x_l)\leq f(x_i)$ for all $i \in \{1,2,\ldots ,n \}$ and $f(x_h)\geq f(x_i)$ for all $i \in \{1,2,\ldots ,n \}$. Clearly such $l$ and $h$ exist. Now if $f(x_l)=f(x_h)$, then all $f(x_i)$ are equal so there is nothing to prove. So $f(x_l)<f(x_h)$ and so now assume WLOG  $x_l<x_h$. Now consider the function $f(x) - \frac{1}{n}(f(x_1)+ \ldots f(x_n))$ in the interval $[x_l,x_h]$. Clearly $f(x_l) \leq 0$ and $f(x_h)\geq 0$, hence by IMVT $\exists c \in [x_l,x_h]$ with $f(c) - \frac{1}{n}(f(x_1)+ \ldots f(x_n)) = 0$.
