Question 1153229: how many rows are in the corner section of a tennis court stadium containing 255 seats if the first row has a 12 seats and the each successive row has 3 additional seats
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
The numbers of seats form an arithmetic sequence:
12, 15, 18, ..., 12+3(n-1)
We need to determine the number of rows (n, the number of terms in the sequence) for which the sum is 255. Algebraically, the sum is
(number of terms) times (average of all the terms)
which for an arithmetic sequence is
(number of terms) times (average of first and last terms)
n = -17 or n = 10; but negative numbers don't make sense for the problem. So n=10.
ANSWER: 10 rows
If a formal algebraic solution is not required, the problem can be solved fairly easily using mental arithmetic.
The sum of 255 is the product of two numbers -- the number of rows, and the average number of seats in each row. That means that twice the sum, 510, is the product of two whole numbers -- the number of rows, and the SUM of the numbers in the first and last rows.
Knowing that the sum of the numbers of seats in the first and last rows is a multiple of 3 (because the number in each row is a multiple of 3), some quick mental arithmetic can find the solution.
Perhaps the most obvious pair of numbers to try is
510 = 10*51
This potential solution requires 10 rows, with 12+9(3) = 39 seats in the last row, making the sum of the numbers of seats in the first and last rows 12+39 = 51 -- and that satisfies the conditions of the problem.
So again the answer is 10 rows.
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