Seating Gallery: 900 Invitations and an 80-Seat Row Limit
A government senior secondary school in Haryana is building a stepped seating gallery for its annual function. In the plan the first row nearest the stage has 24 seats, and every row behind it has 4 more seats than the row in front, continuing that pattern. The ground behind the last row ends at the boundary wall, so the gallery can never exceed 20 rows, and no single row may exceed 80 seats because of the width of the ground. The organising committee has sold 900 invitations on the assumption that every invited guest needs a seat, and the works department has fixed the extra-row spacing at 4 seats. The maths teacher is asked to certify the plan: how many rows are actually needed for 900 seats, whether the 80-seat limit is respected, and whether a gentler pattern of 3 extra seats per row would seat the same 900 guests within the same two limits. The contractor is waiting for a yes or no.
1. How many seats does the originally planned 15-row gallery provide, what is the last row's seat count, and does the plan meet the 900-guest demand?
Read the worked answer
Here a = 24, d = 4, n = 15.
S₁₅ = (15/2)[2(24) + 14(4)] = (15/2)(48 + 56) = (15/2)(104) = 780 seats.
Last row = a₁₅ = 24 + 14(4) = 80 seats.
Verdict: 780 < 900, so 120 invited guests would have no seat - the plan fails on capacity. The last row of exactly 80 is at the structural limit with zero margin, which is acceptable only if no further row is added with the 4-seat spacing.
2. With 4 extra seats per row, find the minimum number of rows needed for 900 seats and check whether the 80-seat row limit allows it.
Read the worked answer
S_n = (n/2)[48 + 4(n − 1)] = (n/2)(44 + 4n) = 2n² + 22n.
Require 2n² + 22n ≥ 900 ⇒ n² + 11n − 450 ≥ 0.
Root: n = [−11 + √(121 + 1800)]/2 = (−11 + √1921)/2 ≈ (−11 + 43.83)/2 ≈ 16.4, so round up to 17.
Check: n = 16 gives 2(256) + 22(16) = 512 + 352 = 864 < 900 (36 short); n = 17 gives 2(289) + 22(17) = 578 + 374 = 952 ≥ 900 ✓.
But the last row at n = 17 is 24 + 16(4) = 88 seats, which exceeds the 80-seat limit. Since 17 is already the smallest adequate n and it breaks the width rule, the 4-seat spacing cannot be certified at any number of rows. Answer to the contractor: no.
3. Test the alternative of 3 extra seats per row: find the minimum number of rows for 900 seats, the last row's seat count, and give a clear recommendation.
Read the worked answer
With d = 3: S_n = (n/2)[48 + 3(n − 1)] = (n/2)(45 + 3n) = 1.5n² + 22.5n.
Require 3n² + 45n − 1800 ≥ 0 ⇒ n² + 15n − 600 ≥ 0.
Root: n = [−15 + √(225 + 2400)]/2 = (−15 + √2625)/2 ≈ (−15 + 51.23)/2 ≈ 18.1, so n = 19.
Check: n = 18 gives (18/2)(45 + 54) = 9 × 99 = 891 < 900 (9 short); n = 19 gives (19/2)(45 + 57) = 9.5 × 102 = 969 ≥ 900 ✓.
Last row = 24 + 18(3) = 78 ≤ 80 ✓, and 19 rows ≤ 20 rows ✓.
Recommendation: reject the 4-seat spacing, adopt 3 extra seats per row with 19 rows. It seats 969, keeps every row inside the 80-seat width limit, stays within the 20-row wall limit, and leaves 69 spare seats for staff and late guests.