The problem describes a situation where the first row of a theater has 15 seats, and each subsequent row has 2 more seats than the row in front of it. There are 11 rows in total, and we need to find out how many seats there are in total.

This situation describes an arithmetic sequence where:

• The first term ($a_1$) is 15.
• The common difference ($d$) is 2.
• The number of terms ($n$) is 11.

The general formula for the $n$-th term of an arithmetic sequence is:

$a_n = a_1 + (n - 1) \times d$

So, the number of seats in the $n$-th row is:

$a_n = 15 + (n - 1) \times 2$

The total number of seats in the theater is the sum of the seats in all rows, which is the sum of the arithmetic sequence. The sum $S_n$ of the first $n$ terms of an arithmetic sequence is given by:

$S_n = \frac{n}{2} \times (a_1 + a_n)$

Let's calculate the sum step by step.

1. Find the number of seats in the 11th row: $a_{11} = 15 + (11 - 1) \times 2 = 15 + 20 = 35$

2. Find the sum of the sequence: $S_{11} = \frac{11}{2} \times (15 + 35) = \frac{11}{2} \times 50 = 11 \times 25 = 275$

So, the total number of seats in the theater is 275.

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. What if the common difference was 3 instead of 2? How many seats would there be?
2. How many seats would there be if there were 15 rows instead of 11?
3. How can you find the total number of seats if the first row had 20 seats instead of 15?
4. What is the total number of seats if the common difference was negative, say $-2$?
5. How does the sum formula change if the common difference $d$ is not constant?

Tip: Understanding how to sum an arithmetic sequence is useful not just in math problems, but also in real-world scenarios involving uniform increases or decreases, such as seating arrangements, budgeting, or even architectural design.