Posted by **Anonymous** on Sunday, November 1, 2009 at 12:19pm.

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

- math- Algebra 2 -
**jim**, Sunday, November 1, 2009 at 12:37pm
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.

- math- Algebra 2 -
**Kaps**, Sunday, November 1, 2009 at 12:37pm
you have 49 rows. row 1 has 40 seats. all u do is 40(n). 49*3=147 + 40= 187

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