A theater has 20 seats in the first row. Each subsequent row has 4 more seats than the row before it. How many seats are in the 10th row?

The seat counts form an arithmetic sequence with first term $20$ and common difference $4$. The $n$th term is $a_n=20+4(n-1)$, so the $10$th row has $20+4(9)=20+36=56$ seats, choice A. Choice D ($60$) uses $n=10$ rather than $n-1=9$ additional steps, which is the $11$th row. Choice B ($44$) uses $6$ steps. Choice C ($40$) uses $5$ steps. The first row is the starting term, so the number of added differences is one less than the row number.