Let $\lim\limits_{n\to\infty}a_{n+1}-a_{n}=\alpha$. Show that $\lim\limits_{n\to\infty} \frac{a_n}{n}=\alpha$.

I'll turn this into a different exercise.

Suppose that $\{x_n\}$ approaches $x$. Then prove that $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^nx_k=x$$ as well.

If you complete this, your current problem can be addressed. Since $\{a_{n+1}-a_n\}$ approaches $\alpha$, we can conclude $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n(a_{k+1}-a_k)=\alpha$$ The sum however is telescoping. Thus $$\lim_{n\rightarrow\infty}\frac{1}{n}(a_{n+1}-a_1)=\alpha.$$ Algebra of Limits finishes this.

HINT: It is easy to observe that $a_n=C+n\alpha+b_n$ for large $n$, where $(b_n)$ is some sequence tending to 0.

Let $b_n = a_n - a_{n-1}$ then $a_n = a_1 + b_1. + b_2 + ....+ b_n$. Apply Cesaro theorem to. $b_n$.