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How Many Seats?

Date: 01/15/2003 at 11:25:35
From: Shay
Subject: Word problem

How do you increase without using powers?

There are 40 rows in an auditorium. Each has two more seats than the 
previous row. There are 1960 seats in all. How many seats are in the 
first row of the auditorium?

Date: 01/18/2003 at 12:16:49
From: Doctor Edwin
Subject: Re: Word problem

Hi, Shay.

We need a formula to that gives us the number of seats, based on the 
number of rows and the number of seats in the starting row. If we call 
the number of seats in the first row X, then we can write, 

  seats = (some formula involving X)

   1960 = (some formula involving X) 

and then solve for X and we're done.

The formula has to be something like:

   1960 = X + (X + 2) + (X + 4) + (X + 6) ... (X + 78)

We can simplify that to:

   1960 = 40X + (2 + 4 + 6 + 8 + ... + 78)

Now you can just use a calculator to add all the even numbers from 2 
to 78, solve the equation for X, and you'll have your answer.

But there's a nice shortcut for adding up a series of numbers that 
increases evenly. Let's call N the number of numbers:

   sum = N  *  (the average of all the numbers)

How do we get the average of all the numbers? We just add them all 
up and... oops. That wasn't much of a shortcut, was it? Fortunately, 
we have another trick. The average of a series of numbers that 
increases evenly is the same as the average of the first and last 
numbers (I'll explain why below). That leaves us with:

              (first number + last number) 
   sum = N  *  ----------------------------

If we substitute in the information from your problem, we get

               (X + (number of seats in last row))
  1960 = 40 *  -----------------------------------

Now, I said that the average of a series of numbers is the same as the 
average of the first and last numbers if they increase evenly. Why is 
that true? Well, think of it this way. If you have a sequential group 
of numbers, the average is going to be in the middle. Let's say that 
you take a number that's 5 higher than the average away from the 
group. The average would go down, right? What if you took away one 
that was 5 higher and one that was 5 lower? The average would stay 
the same. So you could take away matched pairs of numbers from the 
set until you were left with just the first and last ones, without 
changing the average at all. If there is an odd number of numbers in 
the group, you'll be left with one right smack in the middle. Taking 
that one out won't change the average either.

So how do you add up a bunch of integers? Well, let's start simply:

    sum   = (all the integers added up)

Now, if I divide something by a number and then multiply it by the 
same number, I haven't changed it, right? so I can do this:

                                   (all the integers added up)
    sum   = (how many integers) *  ---------------------------
                                       (how many integers)

But "all the integers added up divided by the number of integers" is 
just a way of saying, "the average integer." So I can rewrite my 

    sum   = (how many integers) * (the average integer)

                    (seats in first row + seats in last row)
   seats  =  rows * ----------------------------------------

                    (seats in first row + seats in last row)
   seats  =  rows * ----------------------------------------

We know that there are 40 rows and 1960 seats:

                    (seats in first row + seats in last row)
    1960  =   40  * ----------------------------------------
Let's call the number of seats in the first row F. How many seats are 
there in the last row, 39 rows later? F + something, right?

So we have:

                                (F + (F + something))
    1960  =   40  *             ---------------------

Can you take it from here? Write back if you need more help.

- Doctor Edwin, The Math Forum 
Associated Topics:
Middle School Word Problems

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