jmc

If $a_n=an+b$ and $b_n=cn+d$ are two arithmetic sequences, then their term-wise product takes the form $$a_n{b}_n=(an+b)(cn+d)=A{n}^2+Bn+C,$$where $A,B,C$ are constants. Therefore, some sequence $x_n=A{n}^2+Bn+C$ has $x_0=1440,$ $x_1=1716,$ and $x_2=1848.$ Then we have the equations $$\begin{array}{rl}C& =1440,\\ A+B+C& =1716,\\ 4A+2B+C& =1848.\end{array}$$Subtracting $C=1440$ from the second and third equations, we have $A+B=276$ and $4A+2B=408,$ or $2A+B=204.$ Then $$A=(2A+B)-(A+B)=204-276=-72,$$and so $B=276-A=348.$ We conclude that the given sequence has the general formula $$x_n=-72n^2+348n+1440.$$Then the eighth term is $$\begin{array}{rl}{x}_7& =-72\cdot 7^2+348\cdot 7+1440=\boxed{348}.\end{array}$$