I've try solve this question, but I haven't sucess...
The problem is the following:
A continuous functions $f:[a,b]\rightarrow \mathbb{R}$ assume positive and negative values in its domain, show that there exists $a_1,a_2,\ldots,a_k$, k numbers that are arithmetic progression and it is in the domain such that $$f(a_1)+f(a_2)+\cdots+f(a_k)=0$$
Someone can help me with this question?
Fix $k$. Let the domain of the function include the interval $[a,b]$, where $f(a) \times f(b) < 0 $.
Consider the value of $ f(a) + f( a + \frac{b-a}{k-1}) + \ldots + f(b) $. If this is 0, we are done. Otherwise, WLOG we may assume that it has different sign as $f(a)$.
Now, consider the function
$$F(a, \epsilon) = f(a) + f(a + \epsilon) + f(a + 2 \epsilon) + \ldots + f(a + (k-1)\epsilon ), $$
where $\epsilon \in [0, \frac{b-a} {k-1}] $.
Apply the Intermediate-Value Theorem, and we are done.