Name: David Evaska Who is asking: Parent Level: Middle Question: There are 10 students in the first row of seats in an auditorium 12 in the second 14 in the third and 2 additional in each seat. The total number of rows is 40. I know the answer is 1960 can you please show me the formula step by step. Thanks Hi David, I don't like memorizing formulas and I do so as little as I possibly can. I am going to show you how to solve this problem without using any formulas. Your sequence is 10, 12, 14,... with 40 terms (rows). First, you need to know the last term. When you go from one row to the next you add 2 seats. You do this 39 times to get to the 40th row and hence the 40th row has 10 + 2(39) = 88 seats. Thus you need to add 10 + 12 + 14 +...+ 84 + 86 + 88 Rather than add from the first to the last, start at both ends and add toward the center. 10 + 88 = 98 12 + 86 = 98 14 + 84 = 98   .   .   . Since there are 40 rows, you get this sum of 98 twenty times. Hence 10 + 12 + 14 +...+ 84 + 86 + 88 = 20(98) = 1960 This technique can be described by a formula which is now easy to see. The sum is (The first term plus the last term) times (the number of terms)/2 This even works if the number of terms is odd. If there wer 41 rows then you would have 10 + 12 + 14 +...+ 84 + 86 + 88 + 90 and 10 + 90 = 100 12 + 88 = 100 14 + 86 = 100   .   .   . This time there is a row remaining in the center with 50 seats. Thus 10 + 12 + 14 +...+ 84 + 86 + 88 + 90 = 20(100) + 50 = 2050 You can check that this is still (The first term plus the last term) times (the number of terms)/2 Cheers, Penny Go to Math Central