Terms from the sums
A hall's rows hold 16, 19, 22, 25, 28 seats. The sums so far are 16, 35, 57, 82, 110. These are running totals. Subtract each total from the next: 19, 22, 25, 28. These are the list's terms after the first.
- The running totals are bars: S₁ = 16, S₂ = 35, S₃ = 57, S₄ = 82.
- Bar S₁ has no bar before it. So a₁ = S₁ = 16.
- Cut S₂ level with S₁. The top piece is a₂ = S₂ − S₁ = 35 − 16 = 19.
- Cut S₃ level with S₂: a₃ = 57 − 35 = 22. Cut S₄ level with S₃: a₄ = 82 − 57 = 25.
Rule: a₁ = S₁, since nothing comes before the first term. Then aₙ = Sₙ − Sₙ₋₁ for n = 2, 3, 4, and so on.
Here Sₙ is the total of the first n terms. Sₙ₋₁ is the total of the first n − 1 terms. Sₙ has one more term than Sₙ₋₁. So Sₙ − Sₙ₋₁ is the nth term. As bars, Sₙ₋₁ is the part of bar Sₙ below the cut. The top piece is the nth term.
Sometimes Sₙ is given as a formula. Take Sₙ = 2n² + 3n:
| n | Sₙ = 2n² + 3n | aₙ = Sₙ − Sₙ₋₁ |
|---|---|---|
| 1 | 5 | 5 = S₁ |
| 2 | 14 | 14 − 5 = 9 |
| 3 | 27 | 27 − 14 = 13 |
| 4 | 44 | 44 − 27 = 17 |
The list reads 5, 9, 13, 17. The steps are all 4 so far. Now do the same subtraction with letters. Sₙ₋₁ = 2(n − 1)² + 3(n − 1) = 2n² − 4n + 2 + 3n − 3 = 2n² − n − 1. So aₙ = (2n² + 3n) − (2n² − n − 1) = 4n + 1. The step is 4 for every n. So this Sₙ describes an AP.
Drag each bar apart. Cut it level with the bar before. The top piece is a term.
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