Concert Chair Pattern: Testing a Row Rule and a Chair Count
Chairs for the P5 concert are set out in the school hall in a regular pattern. The front row has 18 chairs, and each row behind it has 4 more chairs than the row directly in front, so row 2 has 22 and row 3 has 26. The hall can hold 25 rows. The concert committee has 1,500 chairs in storage and plans to set up only the first 24 rows, expecting that this will leave a few chairs spare. A parent volunteer is not so sure and wants the figures checked before the chairs are moved. One pupil claims that row 20 will have 98 chairs. The committee also wants to reserve the back rows - the ones with more than 100 chairs - for guests, so it needs to know exactly which row is the first to go over 100 chairs.
1. Write a rule for the number of chairs in row n, and use it to check the pupil's claim that row 20 has 98 chairs.
Read the worked answer
Row 1 = 18, and each row adds 4, so the number of chairs in row n is
18 + 4(n − 1) = 4n + 14
Check against the given rows: n = 1 gives 18 ✓, n = 2 gives 22 ✓, n = 3 gives 26 ✓.
Row 20 = 4 × 20 + 14 = 94 chairs, not 98.
The pupil most likely used 18 + 4n = 98, which forgets that the first row already contains the starting 18 and adds only 3 increases (not 20) before row 20. 98 is actually row 21: 4 × 21 + 14 = 98.
2. Are 1,500 chairs enough to set up the first 24 rows? Support your answer with a total and say what the committee should do.
Read the worked answer
Row 24 = 4 × 24 + 14 = 110 chairs.
Total for 24 rows = (number of rows ÷ 2) × (first + last) = (24 ÷ 2) × (18 + 110) = 12 × 128 = 1,536 chairs.
A useful check is pairing: 18 + 110 = 128, 22 + 106 = 128, and so on - 12 pairs of 128.
1,536 > 1,500, so the committee is 36 chairs short; the plan does not work.
Options: set up 23 rows only (row 23 = 106, total = 11.5 × 124 = 1,426 chairs, leaving 74 spare) or borrow 36 extra chairs. Simply stopping partway through row 24 would leave the last row incomplete.
3. Which row is the first to have more than 100 chairs?
Read the worked answer
Solve 4n + 14 > 100.
4n > 86, so n > 21.5, meaning n = 22.
Check the boundary: row 21 has 4 × 21 + 14 = 98 chairs, which is not more than 100; row 22 has 4 × 22 + 14 = 102 chairs.
So row 22 is the first row over 100 chairs, and rows 22 to 25 (102, 106, 110, 114) are the guest rows. Remember that the inequality is strict: a row of exactly 100 chairs would not count.