Determine the seating capacity of an auditorium with 30 rows of seats if there are 20 seats in the first row, 24 seats in the second row, 28 seats in the third row, and so on

The seating capacity of the auditorium is 2,340.

Step-by-step explanation:

This question relates to an Arithmetic Progression (AP).

An Arithmetic Progression (AP) is a sequence of numbers in which the differences of every two consecutive numbers or terms are constant or the same.

Therefore, the seating capacity can be calculated using the an arithmetic progression (AP) formula as follows:

Seating capacity = (n / 2) * (2a_1 + (n - 1)d) (1)

Where;

n = number of rows = 30

a_1 = Number seats in the first row = 20

d = Common difference = 24 - 20 = 28 - 24 = 4

Substituting the values into equation (1), we have:

Seating capacity = (30 / 2) * ((2 * 20) + ((30 - 1) * 4))

Seating capacity =  15 * (40 + (29 * 4))

Seating capacity =  15 * (40 + 116)

Seating capacity =  15 * 156

Seating capacity = 2,340

Therefore, the seating capacity of the auditorium is 2,340.

step-by-step explanation:

why? because i noticed a pattern. 60 120. those numbers without the zero, make the 6 times tables leaving your answer to be b since 6 times 4 is 24 and adding a zero making it 240.

using pemdas, (order of operations) you'll see the answer is 24.