Find the number of seats in the 20th row using the answer from Part A. Then, use the first and last term in the sum of a finite arithmetic series formula to calculate the total number of seats.
C
c
When both the costs and the number of seats vary, it is a good idea to organize the information in a table.
Here, a is the first term and d is the common difference. In our case, the common difference is d= 3 and the first term is a= 18. Let's substitute these values in the above formula and simplify.
The number is seats in the nth row is given by the above formula. Let's use it to write an arithmetic series in summation notation. Since the theater has 20 rows, the lower and upper limits are 1 and 20, respectively.
∑_(n=1)^(20)(15+3n)
b
We can use the explicit formula obtained in Part A to find the last term in the series. Once we have that, we can find the sum using the Sum of Finite Arithmetic Series formula. Since the theater has 20 rows, the last term of the series can be obtained by substituting n=20 in the formula a_n=15+3n.
Since both the costs and the number of seats vary, it is a good idea to organize the information in a table. In this case, the cost per row will decrease by 5 as the row numbers increase by 5.
Rows
Cost per Seat
1-5
$ 60
6-10
$ 55
11-15
$ 50
16-20
$ 45
To find the total possible revenue for the theater at these prices we need to be explicit about the number of seats in each row. Let's add a column to our table that shows the total number of seats at each price point. To obtain the number of seats of each row, we use the formula a_n=15+3n and substitute the number of the row for n.
Rows
Cost per Seat
Total Seats in Group
1-5
$ 60
18+21+24+27+30 =120
6-10
$ 55
33+36+39+42+45 =195
11-15
$ 50
48+51+54+57+60 =270
16-20
$ 45
63+66+69+72+75 =345
We can now multiply the number of seats in each group of rows by the corresponding cost.
Rows
Cost per Seat
Total Seats in Group
Revenue per Group
1-5
$ 60
120
60* 1202 = $ 7200
6-10
$ 55
195
55* 195 = $ 10 725
11-15
$ 50
270
50* 270 = $ 13 500
16-20
$ 45
345
45* 345 = $ 15 525
Let's find the total revenue by adding the revenue from each group together.
7200+ 10 725+ 13 500+ 15 525= $ 46 950
A full house at these price points can generate a revenue of $46 950.
