Auditorium Seating: Arithmetic Progression Under a Width Constraint
The senior wing of a school is planning its annual cultural programme in a rectangular hall whose interior width is 24.6 m. Seats must be arranged in straight rows parallel to the stage. The fire-safety certificate permits at most 26 rows, and the contractor proposes an arithmetic arrangement: 18 seats in row 1, 22 in row 2, 26 in row 3, and 4 more seats in every following row. Two rules are fixed: a clear aisle of 0.3 m must be left on each side of every row, and each seat needs 0.60 m of row width. The contractor has already printed a plan claiming 1,768 seats in 26 rows. Separately, the principal suggests a gentler arrangement - 18 seats in row 1, increasing by 3 per row, again for 26 rows. Tickets sell at ₹80 each and the printer's deadline is tomorrow, so the committee must decide today how many tickets it can honestly print.
1. Check the contractor's printed figure of 1,768 seats and compute the total for the principal's alternative. Which plan yields more seats, and by how many tickets?
Read the worked answer
Contractor's plan: a = 18, d = 4, n = 26. S = n/2 × [2a + (n − 1)d] = 13 × [36 + 25(4)] = 13 × 136 = 1,768 seats. The printed figure is arithmetically correct.
Principal's plan: a = 18, d = 3, n = 26. S = 13 × [36 + 25(3)] = 13 × 111 = 1,443 seats.
Difference = 1,768 − 1,443 = 325 seats, i.e. ₹26,000 of extra revenue at ₹80 a ticket.
So on paper the contractor wins. But arithmetic correctness is not the same as physical possibility - that check is made in the next task.
2. Using the aisle and seat-width rules, find the last row of the contractor's progression that can actually be seated, and the total seats in the rows that comply.
Read the worked answer
Usable row width = 24.6 − (2 × 0.3) = 24.0 m. Maximum seats per row = 24.0 ÷ 0.60 = 40 seats.
Row n of the contractor's plan has 18 + 4(n − 1) seats, so we need 18 + 4(n − 1) ≤ 40 → 4(n − 1) ≤ 22 → n − 1 ≤ 5.5 → n ≤ 6.5.
Hence only n = 6 rows obey the rule. The rows hold 18, 22, 26, 30, 34, 38 seats, giving S = 6/2 × (18 + 38) = 3 × 56 = 168 seats.
Row 7 would need 42 seats = 25.2 m of seating, more than the 24.0 m available. Row 26 would need 118 seats = 70.8 m, nearly three times the hall's useful width - the printed 26-row plan is impossible well before it reaches the last row.
3. The principal still wants 1,200 seats. Keeping the contractor's 4-seat increment but capping every row at the legal 40 seats, work out the maximum the hall can hold in 26 rows and rule on the 1,200 target.
Read the worked answer
Rows 1–6 as computed: 18 + 22 + 26 + 30 + 34 + 38 = 168 seats. Rows 7–26 = 20 rows, each capped at 40 seats → 20 × 40 = 800 seats. Maximum honest capacity = 168 + 800 = 968 seats.
A quicker ceiling check confirms it: even if every one of the 26 rows were packed to the legal maximum, 26 × 40 = 1,040 seats. Since 1,040 < 1,200, no arrangement whatsoever in this hall can seat 1,200 people within the fire and width rules.
The committee should print 968 tickets (not 1,768), and the 1,200 target can only be met by adding rows beyond the fire limit, widening the hall, or holding two shows. The judgement is that the algebra must be tested against the constraints before money is committed.