Prove that $u_n$ is arithmetic sequence if $\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+\cdots+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$ Let $(u_n)$ be a sequence $u_i\neq0$ and
$$\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+\cdots+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$$ for all $n\geq3$ 
Prove that the sequence $(u_n)$ is arithmetic sequence

You need two key features of an arithmetic progression:
$$a_n = a_1 +d(n-1)$$
and 
$$a_n-a_{n-1}=d$$ (which is a consequence of the previous one).
Thus
$$\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$$
$$d\left(\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n-
1}u_n}\right)=\frac{d(n-1)}{u_1u_n}$$
Now sum $\dfrac{u_1}{u_1 u_n}$ to get
$$\frac{u_1}{u_1 u_n} + d\left(\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}
{u_3u_4}+....+\frac{1}{u_{n-1}u_n}\right)=\frac{u_1+d(n-1)}{u_1u_n}$$
$$\frac{1}{u_n} + \frac{d}{u_1u_2}+\frac{d}{u_2u_3}+\frac{d}{u_3u_4}+....+\frac{d}{u_{n-1}u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$
Now replace $d$ by the differences, conveniently:
$$\frac{1}{u_n} + \frac{u_2-u_1}{u_1u_2}+\frac{u_3-u_2}{u_2u_3}+\frac{u_4-u_3}{u_3u_4}+....+\frac{u_n-u_{n-1}}{u_{n-1}u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$
$$\frac{1}{u_n} + \frac{1}{u_1}-\frac{1}{u_2}+\frac{1}{u_2}-\frac{1}{u_3}+-....+\frac{1}{u_{n-1}}-\frac{1}{u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$
This telescopes, giving
$$\frac{1}{u_1}=\frac{u_1+d(n-1)}{u_1u_n}$$
or
$$u_n=u_1+d(n-1)$$
