o keep the spectators out of the line of flight at an air show, the ushers arranged the show's seats in the shape of an inverted triangle. Kevin, who loves airplanes, arrived very early and was seated in the front row, which contained one seat. The second row contained three seats, and those filled very quickly. The third row contained five seats, which were given to the next five people who came. The next row contained seven seats. This seating pattern continued all the way to the last row, with each row containing two more seats than the previous row. All 20 rows were filled. How many people attended the air sh

Answer:

400 people attended the air show.

Step-by-step explanation:

Arithmetic sequence:

An arithmetic sequence has the following general equation:

[tex]a_n = a_1 + d(n-1)[/tex]

In which [tex]a_n[/tex] is the nth term, [tex]a_1[/tex] is the first term and d is the common difference between the terms.

The sum of the first n terms of an arithmetic sequence is given by:

[tex]S_n = \frac{n(a_1+a_n)}{2}[/tex]

In this question:

First row 1 seat, second 3 seats, 3rd 5 seats and so on.

So we have the following arithmetic sequence:

{1,3,5,7,9,...}

So [tex]a_1 = 1, d = 9 - 7 = 7 - 5 = ... = 2[/tex]

So

[tex]a_n = 1 + 2(n-1)[/tex]

All 20 rows were filled. How many people attended the air show?

Sum of the 20 terms of the progression. So

[tex]S_{20} = \frac{20(1+a_{20})}{2} = 10(1+a_{20})[/tex]

The 20th term is:

[tex]a_{20} = 1 + 2(20-1) = 1 + 2*19 = 1 + 38 = 39[/tex]

So

[tex]S_{20} = 10(1+a_{20}) = 10(1 + 39) = 10(40) = 400[/tex]

400 people attended the air show.

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