Suppose the function $f:\mathbb{R}\to \mathbb{R}$ is continuous. For a natural number $k$, let ${x}_{1},{x}_{2},...,{x}_{k}$ be points in $[a,b]$. Prove there is a point $z$ in $[a,b]$ at which

$f(z)=(f({x}_{1})+f({x}_{2})+...+f({x}_{k}))/k$

So I'm thinking about applying the intermediate value theorem:

If

$a<{x}_{1}<b,a<{x}_{2}<b,...,a<{x}_{k}<b$

then

$f(a)<f({x}_{1})<f(b),...,f(a)<f({x}_{k})<f(b)$

or

$k.f(a)<f({x}_{1})+...+f({x}_{k})<k.f(b)$

$f(a)<(f({x}_{1})+f({x}_{2})+...+f({x}_{k}))/k<f(b)$

But I couldn't think of any way to prove that $f(a)<f({x}_{k})<f(b)$ or is it even true?

EDIT: Thanks everyone for your effort.

$f(z)=(f({x}_{1})+f({x}_{2})+...+f({x}_{k}))/k$

So I'm thinking about applying the intermediate value theorem:

If

$a<{x}_{1}<b,a<{x}_{2}<b,...,a<{x}_{k}<b$

then

$f(a)<f({x}_{1})<f(b),...,f(a)<f({x}_{k})<f(b)$

or

$k.f(a)<f({x}_{1})+...+f({x}_{k})<k.f(b)$

$f(a)<(f({x}_{1})+f({x}_{2})+...+f({x}_{k}))/k<f(b)$

But I couldn't think of any way to prove that $f(a)<f({x}_{k})<f(b)$ or is it even true?

EDIT: Thanks everyone for your effort.