A theater has 45 seats in the first row, 48 in the second, 51 seats in the third row, and so on. How many seats are there in row 15?

This problem involves an arithmetic sequence where each row in the theater increases the number of seats by a constant amount. We can solve this problem by identifying the first term, the common difference, and then finding the term corresponding to the 15th row.

**Step 1: Identify the first term and the common difference**

The first term (\(a_1\)) is the number of seats in the first row:
\[
a_1 = 45
\]

The second term (\(a_2\)) is the number of seats in the second row:
\[
a_2 = 48
\]

The common difference (\(d\)) can be calculated by subtracting the first term from the second term:
\[
d = a_2 - a_1 = 48 - 45 = 3
\]

**Step 2: Find the 15th term**

The general formula for the \(n\)-th term of an arithmetic sequence is:
\[
a_n = a_1 + (n - 1) \cdot d
\]

For the 15th row (\(n = 15\)):
\[
a_{15} = 45 + (15 - 1) \cdot 3
\]

Simplifying further:
\[
a_{15} = 45 + 14 \cdot 3 = 45 + 42 = 87
\]

Therefore, there are **87 seats** in row 15.

Would you like more details or have any questions?

Here are some related questions:
1. How many seats are there in row 20?
2. What is the total number of seats in the first 15 rows?
3. How many rows are needed to reach at least 200 seats in a row?
4. If a new row is added with 5 additional seats than the previous common difference, how many seats are in row 16?
5. How many rows have less than 100 seats?

**Tip:** In an arithmetic sequence, you can quickly find the difference between terms by subtracting consecutive terms, which is useful for identifying patterns and predicting future terms.

Learn how to calculate the number of seats in row 15 of a theater using arithmetic sequences. Starting with 45 seats in the first row, 48 in the second, and increasing by a constant difference of 3 seats per row, this problem walks through the solution step by step. Suitable for students in Grades 6-8.

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