1. Calculate the sum $S=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\ldots+\frac{1}{a_{n} a_{n+1}}$ where $a_{1}, a_{2}, \ldots ., a_{n}, a_{n+1}$ are consecutive terms of an arithmetic progression.
Solution. Setting

$$
a_{2}-a_{1}=a_{3}-a_{2}=\ldots=a_{n+1}-a_{n}=d
$$

and multiplying the given equality by $d$, we have

$$
S d=\frac{d}{a_{1} a_{2}}+\frac{d}{a_{2} a_{3}}+\ldots+\frac{d}{a_{n} a_{n+1}}
$$

i.e.

$$
\begin{aligned}
S d & =\frac{a_{2}-a_{1}}{a_{1} a_{2}}+\frac{a_{3}-a_{2}}{a_{2} a_{3}}+\ldots+\frac{a_{n+1}-a_{n}}{a_{n} a_{n+1}} \\
& =\frac{1}{a_{1}}-\frac{1}{a_{2}}+\frac{1}{a_{2}}-\frac{1}{a_{3}}+\ldots+\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \\
& =\frac{1}{a_{1}}-\frac{1}{a_{n+1}}=\frac{a_{n+1}-a_{1}}{a_{1} a_{n+1}}=\frac{n d}{a_{1} a_{n+1}} .
\end{aligned}
$$