Answer: 348

Solution:

The $n$th term of an arithmetic sequence is linear in $n,$ so the product of corresponding terms of two arithmetic sequences is a quadratic $t_n = an^2 + bn + c.$ Indexing the given terms by $n = 0, 1, 2:$ $$\begin{aligned} c &= 1440, \\ a + b + c &= 1716, \\ 4a + 2b + c &= 1848, \end{aligned}$$ which give $a + b = 276$ and $2a + b = 204,$ so $a = -72,$ $b = 348,$ $c = 1440.$

The eighth term is $t_7 = -72 \cdot 49$ $+ 348 \cdot 7$ $+ 1440 = 348.$ (Indeed $t_n = (180 - 24n)(8 + 3n),$ a product of two arithmetic sequences matching the given terms.)