4.012. The sum of the third and ninth terms of an arithmetic progression is 8. Find the sum of the first 11 terms of this progression.
Solution.

From the condition, we have $a_{3}+a_{9}=8$. Using formula (4.1), we get $a_{1}+2 d+a_{1}+8 d=8, 2 a_{1}+10 d=8$, and using formula (4.5), we find $S_{11}=\frac{2 a_{1}+10 d}{2} \cdot 11=44$.

Answer: 44.