A stadium has 30 rows of seats. The first row has 15 seats and each following row has 4 more seats than the one before. How many seats are there in total?

Step-by-step solution

1. 1

   Model the rows as an arithmetic sequence. Row sizes are $u_1=15$ with common difference $d=4$, and there are $n=30$ rows.

2. 2

   Find the size of the last row. $u_{30}=15+(30-1)\cdot 4=15+116=131\text{ seats}$

3. 3

   Apply the arithmetic-series sum. $S_{30}=\frac{n}{2}\left(u_1+u_n\right)=\frac{30}{2}(15+131)=15\cdot 146=2190$

4. 4

   Cross-check with the other sum formula. $S_{30}=\frac{30}{2}\bigl(2\cdot 15+29\cdot 4\bigr)=15(30+116)=15\cdot 146=2190$ Both forms agree.

5. 5

   Answer. The stadium holds $2190$ seats, option D. The distractor $2250$ is $30\times 75$, the result of using the wrong "average row" of $75$ instead of $73$.