SOLUTION: The seating in a local theatre is arranged so that there are 40 seats in the first row, 42 in the second row , and so on. If there are 25 rows, and the standard ticket price is $1

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Question 1111921: The seating in a local theatre is arranged so that there are 40 seats in the first row, 42
in the second row , and so on. If there are 25 rows, and the standard ticket price is $150
per seat, calculate the total revenue if the theatre is sold out?

Found 2 solutions by amalm06, MathTherapy:
Answer by amalm06(224) About Me  (Show Source):
You can put this solution on YOUR website!
If the theatre is sold out, then all the seats are taken. The number of seats can be found by solving the following series:

sum%2840%2B2i%2Ci=0%2C24%29

The sum of the partial terms of the series is 1600. Therefore, there are 1600 seats in the theatre.

The total revenue is 150*1600=$240000 (Answer)


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
The seating in a local theatre is arranged so that there are 40 seats in the first row, 42
in the second row , and so on. If there are 25 rows, and the standard ticket price is $150
per seat, calculate the total revenue if the theatre is sold out?
This represents the sum of an AP, with:
The 1st term, or a1 being 40
The common difference, or d being 2 (difference in the number of seats, from one row to the next)
n, or number of terms, or number of rows being 25
Sum of an AP formula: matrix%281%2C3%2C+a%5Bn%5D%2C+%22=%22%2C+%28n%2F2%29%282a%5B1%5D+%2B+%28n+-+1%29d%29%29
This becomes: matrix%281%2C3%2C+a%5B25%5D%2C+%22=%22%2C+%2825%2F2%29%282%2840%29+%2B+%2825+-+1%292%29%29 ===> matrix%281%2C3%2C+a%5B25%5D%2C+%22=%22%2C+%2825%2F2%29%2880+%2B+48%29%29 ===> matrix%281%2C3%2C+a%5B25%5D%2C+%22=%22%2C+%2825%2F2%29%28128%29%29 ===> matrix%281%2C3%2C+a%5B25%5D%2C+%22=%22%2C+%2825%2Fcross%282%29%2964%28cross%28128%29%29%29 ===> matrix%281%2C6%2C+a%5B25%5D%2C+%22=%22%2C+25%2864%29%2C+%22=%22%2C+%221%2C600%22%2C+seats%29
With the theater being sold out, 1,600 tickets at $150 per ticket results in a total revenue of: