Calculating the Seating Capacity of an Auditorium

Understanding the seating capacity of an auditorium is crucial for optimal design and functionality. In this article, we will explore how to calculate the total seating capacity of an auditorium that has 21 seats in the first row, with each subsequent row containing one additional seat. We will use mathematical techniques, specifically arithmetic progression, to determine the total number of seats in an auditorium that has 30 rows.

Step-by-Step Calculation

The auditorium layout starts with 21 seats in the first row. Each subsequent row has one more seat than the previous row. This pattern forms an arithmetic progression where the first term (a) is 21, and the common difference (d) is 1.

Finding the Number of Seats in Each Row

The number of seats in the n-th row can be calculated using the formula:
a_{n} a (n - 1)d

Plugging in the values:

a_{n} 21 (n - 1) cdot 1 20 n

Calculating the Total Number of Seats

To find the total number of seats in the auditorium with 30 rows, we need to sum the number of seats in each row. This can be approached using the formula for the sum of an arithmetic series:

S_n frac{n}{2} left[2a (n - 1)dright]

Setting up the equation:

S_{30} frac{30}{2} left[2 cdot 21 (30 - 1) cdot 1right]

Performing the calculations:

S_{30} 15 left[42 29right] 15 cdot 71 1065

Another Approach Using the Sum of the First 30 Natural Numbers

An alternative method involves breaking down the sum of the series. The sum of the first 30 natural numbers (1 to 30) is:

S frac{30 cdot (30 1)}{2} 465

Multiplying this by 20 (the difference between each term and the first term):

20 cdot 465 30 cdot 21 9300 630 1065

Verification Using the Quadratic Formula

To solve the quadratic equation derived, we can use the notion that the sum of the first 30 terms is simplified as follows:

n^2 41n - 2130 0
Using the quadratic formula:

n frac{-41 pm sqrt{41^2 - 4 cdot 1 cdot -2130}}{2 cdot 1}
Simplifying this, we find that the positive solution is:

n 30

Conclusion

The total seating capacity of the auditorium, given 30 rows with an increasing number of seats in each row, is 1065 seats. Understanding and applying arithmetic progressions can help in various design and planning tasks, ensuring efficient utilization of space in auditoriums and other similar venues.

Key Points

First term (a): 21 seats Common difference (d): 1 seat Total number of rows (n): 30 Total seating capacity: 1065 seats

Keywords: Seating Capacity, Arithmetic Progression, Auditorium Design