Question 766248: I really need help solving the following problem. An outdoor amphitheater has 29 seats in the first row, 31 in the second row, 33 in the third row, and so on. There are 36 rows altogether. How many can the amphitheater seat?
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I really need help solving the following problem. An outdoor amphitheater has 29 seats in the first row, 31 in the second row, 33 in the third row, and so on. There are 36 rows altogether. How many can the amphitheater seat?
a(36) = 29+(35)*2 = 99
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S(33) = (33/2)(2+99)
s(33) = 33*50 = 1650 (amphetheater capacity)
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Cheers,
Stan H.
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Answer by MathTherapy(10557)  (Show Source):
You can put this solution on YOUR website! I really need help solving the following problem. An outdoor amphitheater has 29 seats in the first row, 31 in the second row, 33 in the third row, and so on. There are 36 rows altogether. How many can the amphitheater seat?
The last row (36th row) has 99 seats
Formula for the sum of an arithmetic series:  , where:
 is the sum of the terms, or sum of rows, which is sum of 36 rows
n = number of terms (rows) in the series, which is 36
 = 1st term, or amount in 1st row, which is 29
 = nth term, or in this case, the 36th, or last term, which has 99 seats
then becomes:  -----  seat-capacity
You can do a manual check, or use excel to do it!!!
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