1 Answer

Ivan Yurchenko · Answer 1 · 2020-09-02T16:07:58+0000

Instead of using letters from A to AA, let's use numbers. Also, let's number the rows from 0 to 26. So, our labelling will go like this:

$\begin{tabular}c\text{\textbf{Row}}&\text{\textbf{Index}}\\A&0\\B&1\\C&2\\D&3\\\vdots&\vdots\\AA & 26\end{tabular}$

The reason for this labelling is the following: we know that row A has 9 seats, and the following rows have three more seats than the previous one. Let's build a table like the previous one, containing the number of seats for the first few rows:

$\begin{tabular}c\text{\textbf{Index}}&\text{\textbf{\# of seats}}\\0&9\\1&12\\2&15\\3&18\\4 & 21\\\vdots&\vdots\end{tabular}$

So, as you can see, the
-th row has exactly
more seats than the first one.

So, we have the following sequence:

$a_0=9,\ a_1=12,\ a_2=15,\ldots,\ a_n = 3n+9$

Which defines the number of seats for each row.

Answering the questions is now easy: the 27 row, AA, is the 26th term in our sequence:

$a_(26) = 3\cdot 26+9 = 78+9 = 87$

The total number of seats is the sum of all the terms in the sequence:

$\displaystyle \sum_(i=0)^(26)3n+9 = \sum_(i=0)^(26)3n + \sum_(i=0)^(26)9 = 3\sum_(i=0)^(26)n + 27\cdot 9 = 3(26\cdot 27)/(2) + 243 = 3\cdot 13 \cdot 27 + 243 = 1053 + 243 = 1296$

Finally, let
be the number of adult tickets, and
be the number of student tickets. We know that twice as many student tickets were sold as adult tickets, so we have

s = 2a