The first term of an arithmetic sequence is $a_{1}$, the sum of the first and $n$-th term is $s_{a}$, and the corresponding data for another arithmetic sequence are $b_{1}, s_{b}$. A third sequence is formed such that the term of any given order is equal to the product of the terms of the same order from the first two sequences. What is the sum of the first $n$ terms of the third sequence?
Let the $n$-th term of the given sequence be $a_{n}$ and $b_{n}$, their differences be $d_{a}$ and $d_{b}$, respectively, then (assuming $n>1$):

$$
d_{a}=\frac{a_{n}-a_{1}}{n-1}=\frac{s_{a}-2 a_{1}}{n-1}, \quad d_{b}=\frac{s_{b}-2 b_{1}}{n-1}
$$

Based on these, the $i$-th term of the third sequence is

$$
\begin{aligned}
a_{i} b_{i} & =\left[a_{1}+(i-1) d_{a}\right]\left[b_{1}+(i-1) d_{b}\right]= \\
& =a_{1} b_{1}+\left(a_{1} d_{b}+b_{1} d_{a}\right)(i-1)+d_{a} d_{b}(i-1)^{2}
\end{aligned}
$$

Finally, using the known formula for the sum of squares, the desired sum is

$$
\begin{aligned}
S & =a_{1} b_{1}+a_{2} b_{2}+\ldots+a_{i} b_{i}+\ldots+a_{n} b_{n}= \\
& =n a_{1} b_{1}+\left(a_{1} d_{b}+b_{1} d_{a}\right)[0+1+\ldots+(i-1)+\ldots+(n-1)]+ \\
& +d_{a} d_{b}\left[0+1^{2}+\ldots+(i-1)^{2}+\ldots+(n-1)^{2}\right]= \\
& =n a_{1} b_{1}+\frac{a_{1} s_{b}+b_{1} s_{a}-4 a_{1} b_{1}}{n-1} \cdot \frac{(n-1) n}{2}+ \\
& +\frac{\left(s_{a}-2 a_{1}\right)\left(s_{b}-2 b_{1}\right)}{(n-1)^{2}} \cdot \frac{(n-1) n(2 n-1)}{6}= \\
& =\frac{n}{6(n-1)}\left[(2 n-1) s_{a} s_{b}-(n+1)\left(a_{1} s_{b}+b_{1} s_{a}-2 a_{1} b_{1}\right)\right]
\end{aligned}
$$

The last parenthesis can also be written as $a_{1} b_{n}+b_{1} a_{n}$.

Lórincz Éva (Szekszárd, Garay J. Gymnasium IV. grade) Wittmann Róbert (Budapest, Piarist Gymnasium IV. grade)