SOLUTION: The seating in a local theatre is arranged so that there are 40 seat 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

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Question 1111088: The seating in a local theatre is arranged so that there are 40 seat 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 ikleyn, mananth:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
The answer is the product of 150 dollars per seat by the number of seats, which is

the sum of the first 25 terms of an AP with the first term 40 and the common difference 2.


This sum is equal to 


 = 1600 seats,


therefore the revenue is  150*1600 = 240000 dollars.

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On arithmetic progressions,  see the lessons in this site:
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
    - Mathematical induction and arithmetic progressions
    - One characteristic property of arithmetic progressions
    - Solved problems on arithmetic progressions


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The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".


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Be aware:  calculations and the answer by  @mananth  are   W R O N G.



Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
there are 40 seat in the first row,
42 in the second row and so on.
If there are 25 rows
standard ticket price is $150 per seat,
Total number of people when sold out
The seats are in arithmetic sequence
a= 40
d=2
n=25
The total
=
= 25 * 128/2
= 84*25 seats -> 2100
multiply by 150 to get the revenue
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