3. What is the greatest value that the sum of the first $n$ terms of the arithmetic progression $113,109,105, \ldots$ can take?

#
# Solution:

The sum of the first $n$ terms of an arithmetic progression $S_{n}$ takes its maximum value if $a_{n}>0$, and $a_{n+1} \leq 0$. Since $a_{n}=a_{1}+d(n-1)$, from the inequality $113-4(n-1)>0$ we find $n=[117 / 4]=29$.

Then $\max S_{n}=S_{29}=0.5 \cdot(113+113-4 \cdot 28) \cdot 29=1653$.

Answer: 1653.