Question

An auditorium with 52 rows is laid out where 6 seats comprise the first row, 9 seats comprise the second row, 12 seats comprise the third row. The number of seats increases by 3 per row. What is the seating capacity of the auditorium?

4,368

4,160

4,290

1,508

Answer 1

The number of seats per row generate an arithmetic sequence. Let
`a_n` denote the number of seats in the
`n`-th row. We're told that the number of seats increases by 3 per row, so we can describe the number of seats in a given row recursively by



`\begin{cases}a_1=6\\a_n=a_(n-1)+3&\text{for }2\le n\le52\end{cases}`

The total number of seats is given by the summation


`\displaystyle\sum_(n=1)^(52)a_n`

Because
`a_n` is arithmetic, we can easily find an explicit rule for the sequence.


`a_n=a_(n-1)+3`

`a_n=(a_(n-2)+3)+3=a_(n-2)+2\cdot3`

`a_n=(a_(n-3)+3)+2\cdot3=a_(n-3)+3\cdot3`

`\cdots`

`a_n=a_1+(n-1)\cdot3`

So the number of seats in the
`n`-th row is exactly


`a_n=6+(n-1)\cdot3=3+3n`

This means the total number of seats is


`\displaystyle\sum_(n=1)^(52)a_n=\sum_(n=1)^(52)(3+3n)=3\left(\sum_(n=1)^(52)1+\sum_(n=1)^(52)n\right)`

You should be familiar with the remaining sums. We end up with


`\displaystyle\sum_(n=1)^(52)a_n=3\left(52+\frac{52\cdot53}2\right)=4290`