Question

Can anyone solve this this problem in order please

Answer 1

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:

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

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:

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

So, as you can see, the $i$-th row has exactly $3i$ 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:

$\displaystyle \sum_{i=0}^{26}3n+9 = \sum_{i=0}^{26}3n + \sum_{i=0}^{26}9 = 3\sum_{i=0}^{26}n + 27\cdot 9 = 3\dfrac{26\cdot 27}{2} + 243 = 3\cdot 13 \cdot 27 + 243 = 1053 + 243 = 1296$

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

$s = 2a$

Since the show was a sold out, we have

$a+s = 1296 \iff a+2a = 1296 \iff 3a = 1296 \iff a = 432$

So, 432 adult tickets were sold, which implies that

$s = 432\cdot 2 = 864$

student tickets were sold. Let $c$ be the cost of a student ticket. Adult tickets then cost $3c$, and the total revenue is

$432\cdot 3c + 864c = 1296c+864c = 2160c$

So, we have

$2160c =6480 \iff c = \dfrac{6480}{2160} = 3$