The core idea
An arithmetic sequence changes by a constant difference; a geometric sequence changes by a constant ratio. Compound percentage change is geometric because every new percentage uses the updated amount as its base. A model’s constant rate is an assumption, not a promise.
1. An index tells you how many changes occurred
A sequence lists values in an ordered pattern, usually indexed by whole numbers. The index is part of the definition. If a₀ describes today and a₁ describes one week later, a₆ follows six changes. If a₁ is the first week’s delivery, a₆ follows only five changes from that first delivery. A recursive rule calculates the next term from earlier terms; it needs a starting value. An explicit rule gives a term directly from its index. For example, a₀ = 10 and aₙ₊₁ = aₙ+4 generate 10, 14, 18, 22; the explicit rule is aₙ = 10+4n. Write units beside the values and the time interval beside n. A monthly ratio cannot be applied once per day without changing the model.
Sources: OpenStax: Arithmetic Sequences ↗ · OpenStax: Geometric Sequences ↗
2. Constant additions produce arithmetic sequences
With first term a₁ and common difference d, the nth term is aₙ = a₁+(n−1)d. The difference may be positive, zero or negative. To add the first n terms, pair the first with the last, the second with the second-last, and so on. Each pair has the same total a₁+aₙ. Writing the sum forwards and backwards shows 2Sₙ = n(a₁+aₙ), hence Sₙ = n(a₁+aₙ)/2. This argument works even when n is odd because it doubles the whole list before dividing by two. Notice the distinction: aₙ describes one term, while Sₙ counts all n terms. If terms represent weekly additions to a collection, their sum represents the total added; it does not automatically include an earlier stock.
Sources: OpenStax: Arithmetic Sequences ↗ · OpenStax: Series and Their Notations ↗
3. Worked case: a growing weekly delivery
An invented library plan delivers 12 notebooks in week 1 and three more notebooks each following week, for six weeks. These are new deliveries, not the total stock. Here a₁ = 12, d = 3, and n = 6. The sixth delivery is a₆ = 12+5×3 = 27. The list is 12, 15, 18, 21, 24, 27, giving S₆ = 6(12+27)/2 = 117 notebooks. Pairing 12+27, 15+24 and 18+21 independently checks the result: three pairs of 39. If the library initially has 50 notebooks and none leave, its final stock is 167. If some are distributed to learners, subtract those outflows separately. The arithmetic pattern describes the planned inflow only; it cannot by itself describe changing stock when other processes operate.
Sources: OpenStax: Arithmetic Sequences ↗ · OpenStax: Series and Their Notations ↗
4. Constant ratios produce compound change
For a positive starting amount a₀ and fixed positive ratio r, repeated multiplication gives aₙ = a₀rⁿ. A 20% increase means r = 1.20; a 20% decrease means r = 0.80. The next change is calculated from the new base. Increasing and then decreasing by 20% therefore multiplies the original amount by 1.20×0.80 = 0.96, a net decrease of 4%. Equal percentage changes in opposite directions do not cancel. With 0 < r < 1, positive amounts decay towards zero without becoming exactly zero in a finite number of ideal model steps. A practical threshold or rounding rule may end a process earlier. With r > 1, growth accelerates in absolute amount because the same proportion applies to an increasingly large base.
Sources: OpenStax: Geometric Sequences ↗ · OpenStax: Exponential Functions ↗
5. Worked case: storage growth and a threshold
A hypothetical digital archive starts at 200 MB and is projected to grow by 20% at the end of each month. The amounts at months 0, 1, 2 and 3 are 200, 240, 288 and 345.6 MB. A linear prediction adding the initial increase of 40 MB every month would give only 320 MB at month 3. The 25.6 MB difference comes from growth on earlier growth. To find the first complete month when the archive exceeds 400 MB, compare 200(1.2)³ = 345.6 and 200(1.2)⁴ = 414.72. It crosses after four monthly changes. This is conditional planning: deletions, different file types, compression or changing contributions could alter the rate. It is not evidence that a real archive will keep growing at 20%.
A fixed addition and a fixed percentage
| Period n | 100 + 20n | 100 × 1.2ⁿ |
|---|---|---|
| 0 | 100 | 100 |
| 1 | 120 | 120 |
| 2 | 140 | 144 |
| 3 | 160 | 172.8 |
| 4 | 180 | 207.36 |
Sources: OpenStax: Geometric Sequences ↗ · OpenStax: Exponential Functions ↗
6. Add geometric terms and separate mixed processes
For n terms a₁, a₁r, …, a₁rⁿ⁻¹, let their sum be S. Multiply by r and subtract: S−rS = a₁−a₁rⁿ. Thus S = a₁(1−rⁿ)/(1−r) when r ≠ 1; when r = 1, the sum is na₁. This finite formula works for growth and decay. For example, starting with 500 mL and removing 20% of the remaining liquid each cycle removes 100, 80, 64 and 51.2 mL over four cycles. Their sum is 295.2 mL; the remaining amount is 500(0.8)⁴ = 204.8 mL, and both parts add to 500. A rule such as Bₙ₊₁ = 1.2Bₙ+10 combines proportional growth with a fixed addition. It is not a pure geometric sequence; modelling the order of these operations matters.
Sources: OpenStax: Series and Their Notations ↗ · OpenStax: Geometric Sequences ↗
PUT IT INTO PRACTICE
Practice: compare two resource models
- Begin both models at 100 units when n = 0. Model A adds 20 units each period; model B increases by 20% each period. Write explicit rules.
- Calculate both models at n = 1, 2, 3 and 4. Record the amounts in a table with the same units and time intervals.
- Find the first whole period when each model reaches at least 200 units. Explain why one initial matching increase does not establish the same long-term rule.
- Solution: Aₙ = 100+20n gives 120, 140, 160, 180; Bₙ = 100(1.2)ⁿ gives 120, 144, 172.8, 207.36. A reaches 200 at n = 5; B first passes it at n = 4. A keeps a fixed difference; B keeps a fixed ratio. Neither model is justified for a real resource without evidence about its process.
Check your understanding
Why does a₁-based indexing use n−1?
The first term has already been supplied. Reaching the nth term requires only n−1 further changes.
Is the sixth delivery the six-week total?
No. A term and a sum answer different questions. In the notebook example they are 27 and 117 respectively.
Why does a 20% fall not reverse a 20% rise?
The fall uses the increased amount as its base. The combined factor is 0.96, so the original value is not restored.
What happens to the geometric-sum formula at r = 1?
Its written fraction divides by zero, so use the original sum: n identical terms have total na₁.
Can a decreasing geometric model become negative?
Not from a positive start with 0 < r < 1. Every multiplication preserves positivity, even though the values approach zero.
What makes a projection conditional?
It depends on its starting value, period, rate and process remaining applicable. A correct calculation does not verify those assumptions.