## College Algebra (10th Edition)

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The number of seats forms an arithmetic sequence, with $a_{1}=10,d=4,S_{n}=2040$ The sum $S_{n}$ of the first $n$ terms of $\left\{a_{n}\right\}$: $S_{n}=\displaystyle \frac{n}{2}\left[2a_{1}+(n-1)d\right]$ ... which we solve for n $2040=\displaystyle \frac{n}{2}[2(10)+(n-1)4]$ $4080=n[20+4n-4]$ $4080=4n^{2}+16n$ $4n^{2}+16n-4080=0$ $n^{2}+4n-1020=0$ ...use the quadratic formula $n=\displaystyle \frac{-4\pm\sqrt{16-4(1)(-1020)}}{2(1)}$ ... discard the negative solution $n=\displaystyle \frac{-4+64}{2}=30$