Find the first term, last term and the sum of the finite arithmetic sequence.the total seating capacity in an auditorium  is 1,065 the first row has 21 seats and each row has one seat more than the row in front of it. How many row of seats are there in the auditorium 

To solve this problem, we need to identify the first term, last term, and the sum of the finite arithmetic sequence, and then determine the number of rows in the auditorium.

### Given:
- The first term \( a_1 \) (number of seats in the first row) = 21
- The total sum of the sequence \( S_n \) (total seating capacity) = 1,065
- The common difference \( d \) (difference in the number of seats between each row) = 1

### Steps:

1. **Find the number of rows \( n \)**:
   The formula for the sum of the first \( n \) terms of an arithmetic sequence is:
   \[
   S_n = \frac{n}{2} \times (2a_1 + (n-1) \times d)
   \]
   Substituting the given values:
   \[
   1065 = \frac{n}{2} \times (2 \times 21 + (n-1) \times 1)
   \]
   Simplifying:
   \[
   1065 = \frac{n}{2} \times (42 + n - 1)
   \]
   \[
   1065 = \frac{n}{2} \times (n + 41)
   \]
   \[
   2130 = n \times (n + 41)
   \]
   \[
   n^2 + 41n - 2130 = 0
   \]
   Solve this quadratic equation using the quadratic formula:
   \[
   n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \( a = 1 \), \( b = 41 \), and \( c = -2130 \).
   \[
   n = \frac{-41 \pm \sqrt{41^2 + 4 \times 2130}}{2}
   \]
   \[
   n = \frac{-41 \pm \sqrt{1681 + 8520}}{2}
   \]
   \[
   n = \frac{-41 \pm \sqrt{10201}}{2}
   \]
   \[
   n = \frac{-41 \pm 101}{2}
   \]
   Solving for \( n \):
   \[
   n = \frac{60}{2} = 30 \quad \text{(since \( n \) must be positive)}
   \]

2. **Find the last term \( a_n \)**:
   The last term \( a_n \) is given by:
   \[
   a_n = a_1 + (n-1) \times d
   \]
   Substituting the values:
   \[
   a_{30} = 21 + (30-1) \times 1 = 21 + 29 = 50
   \]

### Summary:
- **First term (number of seats in the first row)**: \( a_1 = 21 \)
- **Last term (number of seats in the last row)**: \( a_{30} = 50 \)
- **Sum of the sequence (total seating capacity)**: \( S_n = 1065 \)
- **Number of rows**: \( n = 30 \)

Would you like any further details or explanations on these steps?

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Here are 5 related questions you might find interesting:

1. How do you calculate the common difference if the sequence was geometric?
2. If each row had 2 more seats than the previous one, how would the total capacity change?
3. How can you verify the number of rows using a different method?
4. What if the total seating capacity was 1,500 with the same pattern of seat increase?
5. How does the sum of an arithmetic sequence compare to a geometric sequence?

**Tip:** When solving quadratic equations, always consider both roots, but remember to discard any non-physical (negative) solutions when dealing with real-world scenarios.

Find the first term, last term and the sum of the finite arithmetic sequence. The total seating capacity in an auditorium is 1,065, the first row has 21 seats and each row has one seat more than the row in front of it. How many rows of seats are there in the auditorium?

Explore how to determine the number of rows in an auditorium with a seating capacity of 1,065, where the first row has 21 seats, and each subsequent row has one more seat than the previous. This problem applies key arithmetic sequence concepts and the quadratic formula. Suitable for Grades 9-10.

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