Seating for the Annual Day
A composite school in Belagavi is arranging chairs in its open ground for annual day. The plan has 24 rows facing the stage. The first row has 30 chairs, and each row behind has 4 more chairs than the row in front up to row 20. Rows 21 to 24 are reserved for guests and will have 10 fewer chairs than the normal pattern would give, because a camera platform occupies part of each of those rows. The school has hired 1,150 chairs from a vendor. The fire officer has also insisted that no single row may have more than 110 chairs. Before the chairs are moved, the school wants to know the total seating capacity of the final plan, whether the hired stock is sufficient, which row first breaches the fire limit and by how much, and how many complete rows could be filled if only the 1,150 chairs arrive.
1. Find the total seating capacity of the final 24-row plan and decide whether 1,150 hired chairs are sufficient.
Read the worked answer
The normal pattern is an AP with a = 30 and d = 4. For 24 rows: S24 = 24/2 [2(30) + (24 − 1)4] = 12(60 + 92) = 12 × 152 = 1824.
Rows 21–24 under the normal pattern would be 30 + 20×4 = 110, 114, 118, 122. The plan reduces each by 10, giving 100, 104, 108, 112.
Total = 1824 − 4×10 = 1784 chairs.
Since 1784 > 1150, the hired stock is short by 1784 − 1150 = 634 chairs.
2. Check the fire limit of 110 chairs per row. Which row first breaches it, and what is the minimum reduction needed?
Read the worked answer
Rows 1–20 follow the normal pattern, and the largest is row 20: 30 + 19×4 = 106, which is within 110.
The reserved rows are 100, 104, 108, 112. Only row 24 exceeds 110, by 2 chairs.
Minimum reduction is therefore 2 chairs, bringing row 24 to 110. Note that the unadjusted row 22 would have been 114, but that row is not used at the normal pattern in the final plan.
3. If only the 1,150 hired chairs arrive, how many complete rows can be filled from the front, and what is the shortfall in the next row?
Read the worked answer
For n front rows at the normal pattern, Sn = n/2[60 + (n − 1)4] = n(28 + 2n). S17 = 17(28 + 34) = 17 × 62 = 1054. S18 = 18(28 + 36) = 18 × 64 = 1152.
With 1,150 chairs, 17 rows are complete, using 1,054 chairs; 96 chairs remain. Row 18 needs 30 + 17×4 = 98 chairs, so it is short by 2 chairs. Thus 17 complete rows can be filled, and row 18 is two chairs short.