## Task A-4.4.

The members of the sequence $x_{1}, x_{2}, x_{3}, \ldots$ are obtained by multiplying the corresponding members of two arithmetic sequences. The first three members of the resulting sequence are $x_{1}=1440, x_{2}=1716$ and $x_{3}=1848$. Determine the eighth member of this sequence.
## Solution.

Let $a, a+n, a+2 n, \ldots$ and $b, b+m, b+2 m, \ldots$ be two arithmetic sequences. From the conditions of the problem, we have

$$
\begin{aligned}
a b & =1440 \\
(a+n)(b+m) & =a b+a m+b n+n m=1716 \\
(a+2 n)(b+2 m) & =a b+2 a m+2 b n+4 n m=1848
\end{aligned}
$$

Let's introduce the notations $x=a m+b n$ and $y=n m$.

By substituting the first equation into the other two, we get the system

$$
\begin{array}{r}
x+y=276 \\
2 x+4 y=408
\end{array}
$$

The eighth term of the sequence is therefore

$$
\begin{aligned}
(a+7 n)(b+7 m) & =a b+7(a m+b n)+49 m n & & \\
& =a b+7 x+49 y & & 1 \text { point } \\
& =1440+7 \cdot 276+49 \cdot(-72)=348 . & & 1 \text { point }
\end{aligned}
$$