The Land O’ Lakes High School auditorium has exactly 26 rows of seats. The rows are labeled, in order, from the front of the auditorium to the back from A through Z. There are 8 seats in the row A. Each row after the first row has two more seats than the previous row. There are 10 seats in row B, 12 seats in row C and so on. How many seats are there in row Z?

*The number of seats in each row form the arithmetic sequence 8, 10, 12, 14, 16, …., where a = 8, the common difference is d = 2, and the nth term of the sequence can be determine using the formula a + d(n − 1). Thus, the 26th term of the sequence is 8 + 2(25) = 8 + 50 = 58. Therefore, there are 58 seats in Row Z.*

What is the total number of seats in the Land O’ Lakes High School auditorium?

*From the previous problem, we know that the number of seats in each row form the arithmetic sequence with terms a, a + d, a + 2d, a + 3d, …, a + 24d, a + 25d, where a = 8 and d = 2. It follows that the sum of the terms is a + a + d + a + 2d + a + 3d + ··· + a + 24d + a + 25d = 26a + d(1 + 2 + 3 + ··· + 24 + 25). The sum 1 + 2 + 3 + ··· + 24 + 25 = (1 + 25)(25/2) = 26(25/2) = 13(25) = 325. So, the total number of seats in the auditorium is 26a + 325d = 26(8) + 325(2) = 208 + 650 = 858 seats.*

Land O’ Lakes High School collected $2860 from ticket sales for the winter concert. A ticket was sold for every seat in the auditorium, resulting in a sold out concert. If the price of an adult ticket was three times the price of a student ticket, and twice as many student tickets were sold as adult tickets, what was the price of a student ticket?

*Let x and y represent the number of student and adult tickets sold, respectively. We are told that twice as many student tickets were sold as adult tickets. We also are told that a ticket was sold for every seat, and from the previous problems, we know that there are 858 seats in the auditorium. So we have the equations: x = 2y and x + y = 858. Substituting 2y for x in the second equation, we have 2y + y = 858 → 3y = 858 → y = 286, which means x = 2(286) = 572. So, 286 adult tickets and 572 student tickets were sold for the concert. Now let m and n represent the price of a student and adult ticket, respectively. We are told that a total of $2860 was collected from selling student and adult tickets, and that the price of an adult ticket was three times the price of a student ticket. Based on this information, we can write the following equations: 572m + 286n = 2860 and n = 3m. Substituting 3m for n in the first equation, we have 572m +286(3m) = 2860 → 572m + 858m = 2860 → 1430m = 2860 → m = 2. Therefore, the price of a student ticket was $2.00.*

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