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.

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 $$\frac12(21\cdot 76)=798\;.$$
