Q1.
Exercise: Consider the sequence 1, 4, 7, 10, 13, … Can you predict the next four terms? Can you derive the first 10 terms of the sequence obtained by adding all the terms up to a given term of this sequence? (Hint: The first term is 1. The second term is 1 + 4 = 5, the third term is 1 + 4 + 7 = 12, and so on.)
Answer
The gap between consecutive terms is a constant 3, so the next four terms are 16, 19, 22, 25.
1, 4, 7, 10, 13, 16, 19, 22, 25, 28, …
nth term: tn = 1 + (n – 1) × 3 = 3n – 2
nth term: tn = 1 + (n – 1) × 3 = 3n – 2
Now build the running totals. Take the first ten terms 1, 4, 7, 10, 13, 16, 19, 22, 25, 28 and add them one at a time:
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| term of 1, 4, 7, … | 1 | 4 | 7 | 10 | 13 | 16 | 19 | 22 | 25 | 28 |
| running total Sn | 1 | 5 | 12 | 22 | 35 | 51 | 70 | 92 | 117 | 145 |
So the required sequence is 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.
Why it happens: the gaps of the new sequence are 4, 7, 10, 13, … — the original sequence from its second term on. That is exactly what a running total does: the jump from Sn–1 to Sn is the nth term you just added. Because those gaps themselves increase by a constant 3, the totals grow faster and faster, and the new sequence is not an AP even though the one it came from is.
Did you know? 1, 5, 12, 22, 35, 51, … are the pentagonal numbers. Their nth term is n(3n – 1)/2 — check n = 4: 4 × 11 ÷ 2 = 22.