# Math

An auditorium has 30 rows with 10 seats in the first row, 12 in the second row, 14 in the third row, and so forth. How many seats are in the auditorium

My answer is 1170 but the way I figured out the problem was by listing numbers 1-30. the problem state that the 1st row had 10, 2nd row had 12, and 3rd row had 14 and so forth. So i basically did the same method till i got to 30 and then add up all the numbers which gave me the answer of 1170. Does anybody know a shorter formula i can use?

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1. sum of series, n = 1 to 30
10 + (10+2) + (10+2)+2 etc to 10+29(2)
so this last term is 68
so
Sum = 10 + 12 + 14 ....+64 + 66 + 68
Sum = 68 + 66 + 64 ....+14 + 12 + 10
so
2Sum = 78 + 78 + 78 .... +78 + 78 +78
thirty of those 78s is twice the sum
Sum = 30 *78/2
= 1170

OR LOOK UP ARITHMETIC SERIES in your book or Google it :)

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2. thanks Damon

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3. (n/2)(a1+an)
or
(n/2)[2a1 + (n-1)d]

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4. An auditorium has 30 rows with 10 seats in the first row, 12 in the second row, 14 in the third row, and so forth. How many seats are in the auditorium

My answer is 36, because there as 10 seats in the first row I added it with 14 and 12 because 14 and 12 is equal to 26 if you add a ten thats 36. I'm not the smartest but I think my answer makes sense.

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