8.1.1. (12 points) Among the first hundred elements of the arithmetic progression $3,7,11, \ldots$ find those that are also elements of the arithmetic progression $2,9,16, \ldots$ In your answer, indicate the sum of the found numbers.
Answer: 2870.

Solution.

$$
\begin{gathered}
a_{n}=a_{1}+(n-1) d_{1}=3+(n-1) \cdot 4 \\
b_{m}=b_{1}+(m-1) d_{2}=2+(m-1) \cdot 7 \\
3+(n-1) \cdot 4=2+(m-1) \cdot 7 \\
4(n+1)=7 m, \quad m=4 k, \quad n=7 k-1
\end{gathered}
$$

Consider the sequence of coinciding terms of the progressions $A_{k}$. For $k=1$, we find $n=6$, $m=4$. For $k=2$, we find $n=13$, $m=8$. The first term $A_{1}=a_{6}=23$, the difference $d=a_{13}-a_{6}=$ $51-23=28$. There are a total of $[100: 7]=14$ terms in the sequence, and their sum is

$$
S=\frac{2 \cdot 23+13 \cdot 28}{2} \cdot 15=2870
$$