Auditorium Seating: An AP That Must Also Satisfy the Fire Limit
A high school in Cuttack is finalising the seating plan of its new auditorium. Chairs are arranged in rows, with the number of chairs increasing by a fixed number from row to row. The first plan gives the front row 20 chairs and adds 2 chairs per row, and it is acceptable only if the hall holds exactly 510 chairs. The fire-safety inspector has separately ruled that no single row may contain more than 45 chairs. A second contractor proposes a steeper layout: still 20 chairs in the front row, but 3 extra chairs per row, again totalling exactly 510 chairs, claiming that this arrangement saves floor space. The school must decide which plan can actually be built with whole rows and whole chairs, and whether the inspector's limit is respected. The clerk also needs the number of chairs in the last row of whichever plan is finally approved, so that the furniture order can be placed.
1. Test the first plan: does it give exactly 510 chairs in whole rows, and does it respect the 45-chair row limit?
Read the worked answer
Let there be n rows. The sum of an AP is S = (n/2)[2a + (n − 1)d] with a = 20, d = 2:
(n/2)[40 + 2(n − 1)] = 510 → n[40 + 2n − 2] = 1020 → n(2n + 38) = 1020 → 2n² + 38n − 1020 = 0 → n² + 19n − 510 = 0.
Discriminant = 19² + 4 × 510 = 361 + 2040 = 2401 = 49², so n = (−19 ± 49)/2 = 15 or −34.
n = −34 is rejected (rows cannot be negative), so n = 15 rows.
Last row = a + 14d = 20 + 28 = 48 chairs.
Check: (15/2)(20 + 48) = 15 × 34 = 510 ✓.
But 48 > 45, so the plan violates the fire-safety limit even though it gives the correct total. The correct arithmetic does not make the design legal.
2. Test the second contractor's plan (a = 20, d = 3, total 510).
Read the worked answer
(n/2)[40 + 3(n − 1)] = 510 → n[40 + 3n − 3] = 1020 → n(3n + 37) = 1020 → 3n² + 37n − 1020 = 0.
Discriminant = 37² + 4 × 3 × 1020 = 1369 + 12240 = 13609.
Now 116² = 13456 and 117² = 13689, so 13609 is not a perfect square; √13609 ≈ 116.66.
Hence n = (−37 ± 116.66)/6 ≈ 13.28 or ≈ −25.6 - neither is a positive whole number.
So the second plan is impossible: 13 rows give (13/2)[40 + 36] = 13 × 38 = 494 chairs and 14 rows give (14/2)[40 + 39] = 7 × 79 = 553 chairs. The total of 510 cannot be reached with d = 3. The contractor's claim fails.
3. Keeping 15 rows and exactly 510 chairs, find a whole-number common difference that satisfies the 45-chair limit, and state the first and last rows.
Read the worked answer
With n = 15, the average number of chairs per row is 510 ÷ 15 = 34, and in an AP of 15 terms the average equals the middle (8th) term:
a + 7d = 34 → a = 34 − 7d.
Last row = a + 14d = 34 + 7d, and this must be ≤ 45:
34 + 7d ≤ 45 → 7d ≤ 11 → d ≤ 1.57…
The common difference must be a positive whole number, so d = 1.
Then a = 34 − 7 = 27, last row = 27 + 14 = 41 chairs.
Check: (15/2)(27 + 41) = 15 × 34 = 510 ✓, and every row has at most 41 chairs, so the inspector's limit is met.
The trade-off is worth stating: to keep both the total and the limit, the school must give up the steeper rise of 2 chairs per row.