Let $ f: [a,b]\to \mathbb{R}$ be continuous on $[a,b]$ and $ x_{1},x_{2},...,x_{n} \in [a,b].$Then there is a point $ c \in [a,b]$ such that $f(c)= \frac{f(x_{1})+f(x_{2})+...+f(x_{n})}{n}$.

Can anyone give me some hint which I can use to prove this? Any help would be appreciated. Thanks in advance

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    $\begingroup$ $(f(x_1)+f(x_2)+\dots+f(x_n))/n$ is intermediate between $\min_i f(x_i)$ and $\max_i f(x_i)$.. $\endgroup$ – GReyes May 14 at 20:18
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    $\begingroup$ Oops. Easy it was..thanks. $\endgroup$ – math is fun May 14 at 20:25
  • $\begingroup$ Worth noting: if the $\{x_i\}$ form a partition of $[a,b]$, the case $n\to\infty$ is the First Mean Value Theorem for Integrals. This won't depend on the partition because $f$ is continuous. $\endgroup$ – Integrand May 14 at 21:50

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