3. The sum of fifty natural numbers is 8625. The second number is 5 greater than the first, the third number is 5 greater than the second, the fourth number is 5 greater than the third, ..., the fiftieth number is 5 greater than the forty-ninth number. Determine the largest and smallest addend in the given sum.
Solution. Let $x$ be the first addend. Then the second addend is $x+5$, the third is $x+5+5=x+2 \cdot 5$, the fourth is $x+2 \cdot 5+5=x+3 \cdot 5, \ldots$, the fiftieth addend is $x+49 \cdot 5$. Therefore,

$$
\begin{aligned}
& x+(x+5)+(x+2 \cdot 5)+(x+3 \cdot 5)+\ldots+(x+49 \cdot 5)=8625 \\
& 50 x+(5+2 \cdot 5+3 \cdot 5+\ldots+49 \cdot 5)=8625 \\
& 50 x+5 \cdot(1+2+\ldots+49)=8625 \\
& 10 x+1+2+\ldots+49=1725 \\
& 10 x+\frac{49 \cdot(49+1)}{2}=1725 \\
& 10 x+1225=1725 \\
& 10 x=500 \\
& x=50
\end{aligned}
$$

Thus, the smallest addend is 50, and the largest addend is equal to $50+49 \cdot 5=295$.