Seating Capacity Calculation

The problem asks for the minimum number of rows needed to seat 150 people in an auditorium where seats increase by a fixed amount in each subsequent row. This forms an arithmetic progression.

Auditorium Row Progression

We are given the details of the seating arrangement:

• Number of seats in the first row ($a_1$): 8
• Increase in seats per row ($d$): 4

The total number of seats in the first $n$ rows ($S_n$) can be calculated using the arithmetic progression sum formula:

$ S_n = \frac{n}{2}[2a_1 + (n-1)d] $

Calculating Total Seats

Substitute the given values $a_1 = 8$ and $d = 4$ into the formula:

$ S_n = \frac{n}{2}[2(8) + (n-1)4] $

Simplify the expression:

$ S_n = \frac{n}{2}[16 + 4n - 4] $

$ S_n = \frac{n}{2}[12 + 4n] $

$ S_n = n(6 + 2n) $

$ S_n = 2n^2 + 6n $

Minimum Rows for 150 People

We need to find the minimum integer $n$ such that the total seats $S_n$ is at least 150:

$ S_n \ge 150 $

$ 2n^2 + 6n \ge 150 $

Let's check the total seats for $n=3$ rows (Option C):

$ S_3 = 2(3)^2 + 6(3) $

$ S_3 = 2(9) + 18 $

$ S_3 = 18 + 18 $

$ S_3 = 36 $

According to the provided answer, the minimum number of rows needed is 3.