Answer:
342 seats
Step-by-step explanation:
Given : Theater has 19 rows and the last row has 27 seats. The number of seats in each row increases by 1 as you move toward the back of the section.
To Find: How many seats are in this section of the theater.
Solution:
Since we are given that each row increases by 1 seat
So, we can use arithmetic progression
So, d = 1
No. of rows n = 19
The last row i.e. row 19 has 27 seats
nth term of A.P. = [tex]a_n=a+(n-1)d[/tex]
a is the first term
Substitute n = 19
So, [tex]a_{19}=a+(19-1)(1)[/tex]
[tex]a_{19}=a+18[/tex]
Since row 19 has 27 seats
So, [tex]27=a+18[/tex]
[tex]27 -18=a[/tex]
[tex]9=a[/tex]
Now we are supposed to find the number of seats in the theater.
So, Sum of first n terms in A.P. = [tex]S_n=\frac{n}{2}(a+a_n)[/tex]
Substitute n =19
[tex]S_{19}=\frac{19}{2}(9+27)[/tex]
[tex]S_{19}=342[/tex]
Hence there are 342 seats in the theater.
