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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

Respuesta :

LammettHash
The number of seats per row generate an arithmetic sequence. Let [tex]a_n[/tex] denote the number of seats in the [tex]n[/tex]-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


[tex]\begin{cases}a_1=6\\a_n=a_{n-1}+3&\text{for }2\le n\le52\end{cases}[/tex]

The total number of seats is given by the summation

[tex]\displaystyle\sum_{n=1}^{52}a_n[/tex]

Because [tex]a_n[/tex] is arithmetic, we can easily find an explicit rule for the sequence.

[tex]a_n=a_{n-1}+3[/tex]
[tex]a_n=(a_{n-2}+3)+3=a_{n-2}+2\cdot3[/tex]
[tex]a_n=(a_{n-3}+3)+2\cdot3=a_{n-3}+3\cdot3[/tex]
[tex]\cdots[/tex]
[tex]a_n=a_1+(n-1)\cdot3[/tex]

So the number of seats in the [tex]n[/tex]-th row is exactly

[tex]a_n=6+(n-1)\cdot3=3+3n[/tex]

This means the total number of seats is

[tex]\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)[/tex]

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

[tex]\displaystyle\sum_{n=1}^{52}a_n=3\left(52+\dfrac{52\cdot53}2\right)=4290[/tex]

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