math- Algebra 2
posted by Anonymous on .
An auditorium has 49 rows of seats. The first row contains 40 seats. As you move to the rear of the auditorium, each row has 3 more seats than the previous row.
* I am pretty sure it involves both arithmetic series and explicit formulas but im not sure exactly how to go about it
OK. First, what exactly is the question? How many seats in the last row? How many seats in the auditorium?
Let's figure a model to start.
Row 1: 40
Row 2: 43
Row 3: 46
Row 4: 49
and so on.
We want a formula that, given the row, will tell us how many seats.
Multiplying the row by 3 has to be part of the answer, since there are 3 extra in each row, but that doesn't account for the initial row.
We _could_ subtract 3 from the 40, imagining a row 0 with 37 seats that doesn't exist. Then our formula would be nice and neat:
Seats in row = 37 + row * 3
Or we could equally well start at 40 but subtract 1 from the number of rows, to get:
Seats in row = 40 + (row - 1) * 3
I'll use the first.
s = 3r + 37
So in row 49 we have 3*49 + 37 seats.
you have 49 rows. row 1 has 40 seats. all u do is 40(n). 49*3=147 + 40= 187