The seats represent an arithmetic sequence
the first term is 11 and the common difference = 3
let the first term = a and the common difference = d and the general term = x
the general term = x = a + d(n-1)
at the the row number 40,
the number of seats at row 40 = 11 + 3 (40 - 1) = 128
To find the total seats, we need to find the number of seats at the first and the last rows
the sum of the arithmetic sequence = n/2 * (first term + last term)
so, the number of seats = (40/2) * ( 11 + 128 ) = 20 * 139 = 2780 seats
total seats = 2780