Answer:
1320 seats.
Step-by-step explanation:
The given scenario can be modeled as an arithmetic series.
[tex]\boxed{\begin{minipage}{7.3 cm}\underline{Sum of the first $n$ terms of an arithmetic series}\\\\$S_n=\dfrac{1}{2}n[2a+(n-1)d]$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\ \phantom{ww}$\bullet$ $d$ is the common difference.\\ \phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}[/tex]
The first term is the number of seats in the front row.
Given that the number of seats in a row increases by 2 with each successive row, the common difference is 2.
The nth term is 30 since there are 30 rows in the theater.
Therefore:
Substitute the values into the arithmetic series formula and solve:
[tex]\implies S_{30}=\dfrac{1}{2}(30)[2(15)+(30-1)2][/tex]
[tex]\implies S_{30}=15[30+(29)2][/tex]
[tex]\implies S_{30}=15[30+58][/tex]
[tex]\implies S_{30}=15[88][/tex]
[tex]\implies S_{30}=1320[/tex]
Therefore, the total seating capacity of the theater is 1320 seats.
