Question 178126
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The first row has 10 seats, the second row, 11, and so on; so the 20th row must have 29 seats.


In each of the rows with an even number of seats, the maximum number of students is simply half of that number of seats, so the first row can seat 5, the third row can seat 6, and so on.  So let's first deal with the number of students that can be seated in the rows with an even number of seats.  First calculate the number of seats in these rows:


The sum of any series of integers is given by:


*[tex \Large \sum_{i=1}^n {x_i} = \frac {(a + l)n}{2}]


where <i>a</i> is the first number in the series, <i>l</i> is the last number in the series, and <i>n</i> is the number of elements in the series.


For the rows with an even number of seats, *[tex \Large a = 10], *[tex \Large b = 28], and, since there are an even number of rows, namely 20, exactly half of them or 10 must be rows with an even number of seats so *[tex \Large n = 10].


*[tex \Large \sum_{i=1}^{10} {x_i} = \frac {(10 + 28)10}{2} = \frac {380}{2} = 190]


But only half of these seats can be occupied, so the rows with an even number of seats can accomodate *[tex \Large {{190 \over {2}} = 95}] students.


Now for the rows with an odd number of seats.  Depending on whether the first seat is occupied or not, the 2nd row, containing 11 seats can accomodate either 6 (first seat occupied) or 5 (first seat not occupied) students.  Since the question is asking for maximum capacity, we need to consider the situation where the first seat is occupied for each of these rows.


However, just adding up the number of seats in these rows and then dividing by 2 is insufficient because, in effect, we would be dividing the odd number of seats in each row by 2, giving us a half person for each of these rows, making our overall count *[tex \Large 10 \div 2 = 5] students short.  Therefore:


*[tex \Large \sum_{i=1}^{10} {x_i} = \frac {(11 + 29)10}{2} = \frac {400}{2} = 200]


Now we divide this by 2 and add the extra 5 that this method undercounts:


*[tex \Large (200 \div 2) + 5 = 105]


Add this to the 95 that we obtained for the even rows to get the final tally:


*[tex \Large 105 + 95 = 200]

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