$\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c , c > 0$ . Then, is $\{x_n/n\}$ convergent?

We claim that $x_n/n\rightarrow c$.

As in the comment we write $a_n=x_n-x_{n-1}$. Then $x_n=\sum_{k=1}^n a_k$ $\color{red} { (1) }$. The assumption becomes $a_n\rightarrow c$ and $x_n/n=\frac{1}{n}\sum_{k=1}^n a_k$.

Then

$$x_n/n-c=\left(\frac{1}{n}\sum_{k=1}^n a_k\right) - c = \frac{1}{n} \sum_{k=1}^n (a_k-c)\quad \color{red} { (2) } $$

Let $\epsilon>0$ and choose $N$ large enough such that $|a_n-c|<\epsilon$ for all $n\ge N$. Then

$$\left|\frac{x_n}{n}-c\right|\le \frac{1}{n}\sum_{k=1}^N |a_k-c| + \frac{n-N}{n}\epsilon$$

for $n\ge N$. Now let $n\rightarrow\infty$ while keeping $N,\epsilon$ fixed. Then we obtain

$$\limsup_{n\rightarrow\infty}\left|\frac{x_n}{n}-c\right|\le \epsilon$$

Since $\epsilon$ was arbitrary, the conclusion follows.

Note:

In essence we just proved that a summable series is also Cesàro summable.

That can be derived directly from Stolz-Cesaro theorem (more complete account here) by taking $b_n = n.$

$$\lim_{n \to \infty} x_{n+1}-x_n = \lim_{n \to \infty} \frac{x_n}{n}$$

when $\lim_{n \to \infty} x_{n+1}-x_n$ exists.