Auditorium Rows and Extra Seats
A school auditorium has rows of seats arranged in arithmetic progression. The first row has 18 seats, and each row behind has 3 more seats than the row immediately in front. The caretaker counts 585 seats in the hall. For the annual function, the principal wants at least 620 seats. The only possible change is to add one more row at the back, continuing the same pattern; no seats can be removed or rearranged. A teacher claims that 15 rows are impossible because the total number of seats would then be odd. The fire safety rule also says the last row must not have more than 65 seats. Determine the present number of rows and whether the proposed single extra row is enough; check both the teacher's claim and the safety rule.
1. Find the present number of rows and the number of seats in the last row. Evaluate the teacher's claim about 15 rows.
Read the worked answer
Here a = 18 and d = 3. The sum of n rows is S_n = n/2[2a + (n - 1)d] = n/2[36 + 3(n - 1)] = n/2(3n + 33) = 3n(n + 11)/2. Set this equal to 585: 3n(n + 11)/2 = 585, so n(n + 11) = 390. Thus n² + 11n - 390 = 0, which factorises as (n + 26)(n - 15) = 0. Since n must be positive, n = 15 rows. The last row has 18 + (15 - 1)3 = 18 + 42 = 60 seats. The teacher's claim is false: 585 is odd, but 15 rows are exactly possible. Oddness of the total is not a reason to reject the row count.
2. If one row is added at the back, what is the new total number of seats and the new last-row count? Is the principal's target of at least 620 seats met, and is the fire safety rule satisfied?
Read the worked answer
The next row has 18 + 15 × 3 = 63 seats. Adding it gives 585 + 63 = 648 seats. Alternatively, using n = 16 in the sum formula gives 3(16)(27)/2 = 648. The new last row has 63 seats, which is not more than 65, so the fire safety rule is satisfied. The principal wants at least 620 seats; 648 is greater than 620, so the target is met by 28 seats. One extra row is therefore enough for the target and is allowed under the safety rule.
3. What is the maximum number of extra rows that can be added while keeping the last row at 65 seats or fewer? Justify.
Read the worked answer
The 16th row already has 63 seats. The 17th row would have 63 + 3 = 66 seats, which exceeds the limit of 65. Since the number of seats per row increases by 3 each time, every later row would be even larger. Therefore only the 16th row can be added safely; the maximum number of extra rows is 1. The total then becomes 648 seats, and the last row has 63 seats, so both the safety rule and the principal's 620-seat target are satisfied.