# Summing numbers which increase by a fixed amount (arithmetic progression)

An auditorium has 21 rows of seats. The first row has 18 seats, and each succeeding row has two more seats than the previous row. How many seats are there in the auditorium?

Now I supposed you could use sigma notation since this kind of problem reminds me of it, but I have little experience using it so I'm not sure.

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You may wish to try a more informative title in the future... –  Antonio Vargas Apr 6 '12 at 19:35
How many numbers are you adding up? What is the average of the numbers you are adding? If you can answer both of these questions, what is the sum of the numbers? –  robjohn Apr 6 '12 at 20:35

You could write the total using sigma notation as $$\sum_{k=0}^{20}(18+2k)\,$$ among many other ways, but I’m pretty sure that what’s wanted here is the actual total. You can add everything up by hand, which is a bit tedious, or you can use the standard formula for the sum of an arithmetic progression, if you know it, or you can be clever and arrange the sizes of the rows like this:

$$\begin{array}{} 18&20&22&24&26&28&30&32&34&36&38\\ 58&56&54&52&50&48&46&44&42&40\\ \hline 76&76&76&76&76&76&76&76&76&76&38 \end{array}$$

The bottom row is the sum of the top two, so adding it up gives you the total number of seats. And that’s easy: it’s $10\cdot 76+38=760+38=798$. This calculation is actually an adaptation to this particular problem of the usual derivation of the formula for the sum of an arithmetic progression, which in this particular case looks like this:

$$\begin{array}{} 18&20&22&24&26&\dots&50&52&54&56&58\\ 58&56&54&52&50&\dots&26&24&22&20&18\\ \hline 76&76&76&76&76&\dots&76&76&76&76&76 \end{array}$$

The top row is the original set of row sizes; the second consists of the same numbers in reverse order; and the bottom row is again the sum of the top two. That now counts each seat twice, so the total number of seats is $$\frac12(21\cdot 76)=798\;.$$

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+1, This is along the lines of what I intended to write. –  user21436 Apr 6 '12 at 19:32

If the auditorium is as described, then the first row has $18$ seats, the second row has $20$ seats, the third row has $22$, the fourth row has $24$, and so on. You can see that the $21$-th row has $18+(2)(20)$ seats, that is, $58$ seats. Now you can certainly add up the even numbers from $18$ to $58$. It takes a while.

But the person who described the auditorium lied. Actually, it has the same number of seats in every row, $18+58$, and it is full. In the first row, the first $18$ people are boys, the remaining $58$ are girls. In the second row, the first $20$ are boys, the rest are girls. In the third row, the first $22$ are boys, the rest are girls. This pattern goes on until the very back, where the first $58$ are boys and the remaining $18$ are girls. Perhaps you can make a sketch of the auditorium, and the seating pattern.

Now can you tell me quickly how many boys there are?

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:) Can't resist but give +1. –  user21436 Apr 6 '12 at 19:39

$$s_1 = 18 = 18 + 2(1-1)\\ s_2 = 20 = 18 + 2(2-1) \\ s_3 = 22 = 18 + 2(3-1) \\ \ldots \\ s_{i} = 18 + 2(i - 1) \\ \ldots \\ s_{21} = 18 + 2(21 - 1)$$

Sum:

$$\sum_{i=1}^{21} (18 + 2(i-1))$$

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You know how to take it from here. Don't you? –  user2468 Apr 6 '12 at 19:29

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+1. What's the software you used for plotting? –  user2468 Apr 6 '12 at 21:18
I am interested in JD's question too. :) –  user21436 Apr 6 '12 at 22:59
Nothing fancy. Just MS Paint, and a lot of copying/pasting! Made 1 dot, then 2 then 4, then 8..., then erased along a diagonal, and added some text. –  Nick Alger Apr 6 '12 at 23:32
Very pretty! $\;\;$ –  Brian M. Scott Apr 7 '12 at 12:28

Hint

Arithmetic progression; What is the first term? What is the common difference? What is the sum of first $n$ terms of an arithmetic progression?

For the formula for sum, the "Gaussian trick" works well.

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Wikilink for OP's benifit. –  user2468 Apr 6 '12 at 19:17
This isn't what we're learning in class. My teacher said we can do it if we like but to skip if we're not familiar. So can you spell it out for me? –  David Apr 6 '12 at 19:18
David, maybe you'd like to add what you are learning in class... –  The Chaz 2.0 Apr 6 '12 at 19:23

The $i$-th row has $18+2i$ seats, for $i=0,...,20$. The number of seats is thus $$\sum_{i=0}^{20}18+2i=18\times21+\sum_{i=0}^{20}2i=378+2\frac{20\times 21}{2}=378+420=798$$

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You have a 0-th row... –  Thomas Apr 6 '12 at 19:28
I know, and the answer is correct and all, but with the way the question is worded, we have 1st row to 21st row. (just a tiny thing that would improve the answer IMO) –  Thomas Apr 6 '12 at 19:41