3. In the arithmetic progression $\left(a_{n}\right) a_{1000}=150, d=0.5$.

Calculate: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cdot a_{1581}}+\frac{1}{a_{1581} \cdot a_{1582}}+\ldots+\frac{1}{a_{2019} \cdot a_{2020}}\right)$.
Solution: The expression in parentheses consists of several terms of the form $\frac{1}{x \cdot(x+d)}$, which can be decomposed into the sum of simpler fractions: $\frac{1}{x \cdot(x+d)}=\frac{1}{d}\left(\frac{1}{x}-\frac{1}{x+d}\right)$. Transform the original expression: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cdot a_{1581}}+\frac{1}{a_{1581} \cdot a_{1582}}+\ldots+\frac{1}{a_{2019} \cdot a_{2020}}\right)=$

$$
=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{1}{a_{1580}}-\frac{1}{a_{1581}}+\frac{1}{a_{1581}}-\frac{1}{a_{1582}}+\ldots+\frac{1}{a_{2019}}-\frac{1}{a_{2020}}\right)=
$$

$$
\begin{gathered}
=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{1}{a_{1580}}-\frac{1}{a_{2020}}\right)=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{a_{2020}-a_{1580}}{a_{2020} \cdot a_{1580}}\right)=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{(2019-1579) d}{a_{2020} \cdot a_{1580}}\right)= \\
=99 \cdot 100 \cdot \frac{1}{d}\left(\frac{(2019-1579) d}{a_{2020} \cdot a_{1580}}\right)=99 \cdot 100 \cdot \frac{1}{d} \cdot \frac{440 d}{660 \cdot 440}=15
\end{gathered}
$$

since $a_{2020}=a_{1000}+1020 d=150+1020 \cdot 0.5=660$, and $a_{1580}=a_{1000}+580 d=150+580 \cdot 0.5=440$.

## Answer: 15.