Showing that for all $n\ge1$, there exists $α\in\mathbb{R}^+$ and $c\in(a,b)$ so that $\sum\limits_{i=0}^nf(c+iα)=(n+1)(c+\frac n2α)$ 
Let $f:[a,b]\to (a,b)$ be a continuous function. Show that for all $n\ge 1,\exists \alpha \in\mathbb{R}^+$ and $c\in (a,b)$ so that $\sum\limits_{i=0}^n f(c+i\alpha)= (n+1)(c+\dfrac{n}2 \alpha)$.

Note that $c+\dfrac{n}2\alpha = \dfrac{1}{n+1}\sum\limits_{i=0}^n (c+i\alpha)$. This might not just be a coincidence.
Let $h(x)=f(x)-x.$ Then by continuity $h$ is negative in a neighbourhood around b and positive in a neighbourhood around a. So by the IVT there necessarily exists a point $x\in (a,b)$ where $h(x) = 0.$ That is, $x$ is a fixed point of $f(x)$.
Note that a continuous function $f:[a,b]\to[a,b]$ necessarily has a fixed point (I believe this is known, and the proof is significantly harder than the proof that $h$ has a zero). I think a proof by contradiction might work for this problem. One way to get a contradiction might be to construct a sequence $(x_n)\subseteq [a,b]$ with $x_n\to a$ but $f(x_n)\not\to f(a)$.
 A: Use IVT to approach the problem
Consider $\alpha=0$. Then the problem is to find a root of $f(x)=x$. It can be found since $f(a)>a$ and $f(b)<b$. To obtain $\alpha>0$, we can perturb a bit.
A full proof
Consider $h(x)=f(x)-x$.
Since $h(a) > 0 $ and $h$ is continuous, $h(x)>0$ for $x$ near $a$.
Since $h(b) < 0 $ and $h$ is continuous, $h(x)<0$ for $x$ near $b$.
So we can choose $a<\overline a<\underline b<b$ such that $h(\overline a)>0>h(\underline b)$.
Since $h(x)$ is continuous, $h(x)>0$ for $x$ near $\overline a$.
If $\beta>0$ is small enough, we have $\overline a+i\beta$ is near to $\overline a$ for all $0\le i\le n$ and hence $\sum_{i=0}^nh(\overline a+i\beta)>0$.
Similarly, if $\beta>0$ is small enough, $\sum_{i=0}^nh(\underline b+i\beta)<0$.
Let $\alpha>0$ be small enough so that $\sum_{i=0}^nh(\overline a+i\alpha)>0$ and $\sum_{i=0}^nh(\underline b+i\alpha)<0$.
Note that $h(x+i\alpha)$ is well-defined for $x\in (\overline a, \underline b)$ since $\overline a +i\alpha<c+i\alpha<\underline b+i\alpha$. By IVT, there exist $c\in(\overline a, \underline b)$ such that $\sum_{i=0}^nh(c+i\alpha)=0$. As observed in the question, that means
$$\sum_{i=0}^n f(c+i\alpha)= (n+1)(c+\dfrac{n}2 \alpha)$$
A slight generalization
As the asker observed, $c+\dfrac{n}2\alpha = \dfrac{1}{n+1}\sum\limits_{i=0}^n (c+i\alpha)$ is not a coincidence. The proof above proves, in fact, the following generalization.
Let $f:[a,b]\to (a,b)$ be a continuous function. $d_0+\cdots+ d_n=s$. Then there exists $\alpha \in\mathbb{R}^+$ and $c\in (a,b)$ so that $\sum\limits_{i=0}^n f(c+d_i\alpha)= (n+1)c+s\alpha$.
A: $\def\paren#1{\left(#1\right)}$This problem is better illustrated under a two-dimensional setting. First, the problem can be rephrased as below giving more symmetry:

Let $f \in C([a, b];\, (a, b))$. Prove that for any $n \in \mathbb{N}_+$, there exist $c, d \in (a, b)$ such that$$
\frac{1}{n + 1} \sum_{k = 0}^n f\paren{ \frac{n - k}{n} c + \frac{k}{n} d } = \frac{1}{2} (c + d).
$$

Proof: Define $F(c, d) = \dfrac{1}{n + 1} \sum\limits_{k = 0}^n f\paren{ \dfrac{n - k}{n} c + \dfrac{k}{n} d } - \dfrac{1}{2} (c + d)$, then $F \in C([a, b]^2)$ and$$
F(a, a) = f(a) - a > 0 > f(b) - b = F(b, b).
$$
Therefore, there exist $δ_1, δ_2 > 0$ with$$
F(a, a + δ_1) > 0 > F(b - δ_2, b).
$$
Since the closed triangular region $D$ with vertices at $(a, a + δ_1)$, $(b - δ_2, b)$, $(a, b)$ is connected and above the line $y = x$, there exists $(c, d) \in D$ with $c < d$ and $F(c, d) = 0$.
