3. What is the smallest value that the sum $S_{n}$ of the first $n$ terms of an arithmetic progression can take, given that the sum $S_{3}=-141$ and the sum $S_{35}=35$?
# Solution:

If $a-$ is the first term and $d-$ is the common difference of the arithmetic progression,

$$
\left\{\begin{array} { l } 
{ \frac { a + a + 2 d } { 2 } \cdot 3 = - 141 , } \\
{ \frac { a + a + 34 d } { 2 } \cdot 35 = 35 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a+d=-47, \\
a+17 d=1
\end{array} \Rightarrow 16 d=48 ; d=3, a=-50\right.\right.
$$

The sum $S_{n}$ of the first $n$ terms of the arithmetic progression takes the minimum value if $a_{n}<0$, and $a_{n+1} \geq 0$. Since $a_{n}=a+d(n-1)$, from the inequality $-50+3(n-1)<0$ we find $n=[53 / 3]=[172 / 3]=17$. Then $\min S_{n}=S_{17}=0.5 \cdot(-50-50+3 \cdot 16) \cdot 17=-442$. -442.