Question

A school auditorium has to be constructed to accommodate at least 1500 people. The chairs are to be placed in a concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

Based on the above information, answer the following questions.

1. If the first circular row has 30 seats then how many seats will there be in the 10th row?
2. For 1500 seats, how many rows must the auditorium have?
3. If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after the 10th row?
4. If there were 17 rows in the auditorium, how many seats would be in the middle row?

Answer 1

The seating arrangement follows an Arithmetic Progression (AP):

First term (seats in first row), a = 30

Common difference, d = 10

i. Seats in the 10th row:

Formula: \[T_n = a + (n - 1)d\]

Substitution: \[T_{10} = 30 + (10 - 1) \times 10\]

\[T_{10} = 30 + 90 = 120\]

There are 120 seats in the 10th row.

ii. Number of rows required for 1500 seats:

Given: Total sum \[S_n = 1500\]

Formula: \[S_n = \frac{n}{2}[2a + (n - 1)d]\]

Calculation: \[1500 = \frac{n}{2}[2(30) + (n - 1)10]\]

\[3000 = n[60 + 10n - 10]\]

\[3000 = n(10n + 50)\]

\[10n^2 + 50n - 3000 = 0\]

Dividing the entire equation by 10: \[n^2 + 5n - 300 = 0\]

Factoring: \[n^2 + 20n - 15n - 300 = 0\] 

\[n(n + 20) - 15(n + 20) = 0\] 

\[(n + 20)(n - 15) = 0\] 

\[n = -20 \quad \text{or} \quad n = 15\] 

Since rows cannot be negative, n = 15.

The auditorium must have 15 rows.

iii. Seats still left to be put after the 10th row:

Total seats accommodated up to 10th row:

\[S_{10} = \frac{10}{2}[2(30) + (10 - 1)10]\] 

\[S_{10} = 5 \times [60 + 90]\]

\[= 5 \times 150\]

\[= 750\]

Seats left to be put: \[\text{Remaining seats} = 1500 - S_{10}\]

\[= 1500 - 750\]

\[= 750\]

750 seats are still left to be put.

iv. Number of seats in the middle row if there are 17 rows:

Middle row determination: For n = 17, the middle row is the:

\[\left(\frac{17 + 1}{2}\right)\text{th row} = 9\text{th row}\]

Seats in the 9th row: \[T_9 = a + (9 - 1)d\]

\[T_9 = 30 + 8(10)\]

\[= 30 + 80\]

\[= 110\]

There are 110 seats in the middle row.