Saving, compounding, paying off a loan — and the bridge to vectors and limits.
Every sequence in this stage has been quietly waiting for a job. A row of dots that climbs by a fixed step is a savings plan; a row that multiplies by a fixed ratio is compound interest, a growing population, or a decaying isotope; a balance reborn from last month's balance is a loan. And one famous run of dots — Fibonacci's 1, 1, 2, 3, 5, 8, … — hides a number we are about to meet again and again: the ratio of neighbours creeps toward the golden ratio φ ≈ 1.618 without ever landing on it. That word, creeps toward, is the seed of calculus, and it is how this lesson hands the next strand two gifts: a number that is about to gain a direction (a vector), and an idea that is about to become a limit.
Some growth is patient and flat. You drop $50 into a jar every month; a worker gets a fixed $2000 raise each year; a savings account paying simple interest adds the same dollar amount every period. In each case the difference between consecutive terms is a constant d — an arithmetic model.
Write the deposit each month as a term. With a₁ = 50 after the first month and a step of d = 50, the balance after n months is the n-th term of an arithmetic sequence,
aₙ = a₁ + (n−1)d, and the running total after n periods is the sum Sₙ = n(a₁ + aₙ)2.
Simple interest is arithmetic. Lend a principal P at simple rate r per year and each year it earns the same P·r dollars. After n years the value is P + P·r·n — a first term P and a constant step P·r, so the yearly values 1050, 1100, 1150, … (for P = $1000, r = 5%) fall on a perfectly straight line. The dots are collinear because aₙ is linear in n (Stage 21, linear functions — here d is the slope).
Save $50 a month, starting this month. After n = 12 months, how much sits in the jar? The deposits themselves are constant, so the balance is the running total: S₁₂ = 12·(50 + 600)2 = 6·650 = $3900. (Here a₁₂ = 50 + 11·50 = 600 is the twelfth deposit, not the balance — keep aₙ and Sₙ apart.)
Simple interest adds the same dollars each year; compound interest (next section) multiplies by the same factor. Mixing them up is the most expensive mistake in this whole lesson. And a model lives only at the whole periods n ∈ ℕ — the connecting line is a guide for the eye, not a value you can read between the dots.
Now let each period scale the previous one by the same ratio q. Money left to compound earns interest on its interest, so each year multiplies the balance by 1 + r; a population that grows a fixed percent a year multiplies by the same factor; a radioactive sample loses half its mass each half-life, multiplying by ½. These are geometric models, and the dots curve like an exponential.
A = P(1 + r)n (compound interest — geometric, ratio q = 1 + r)
With P = $1000 and r = 5%, the balance is a₁ = 1000 and ratio q = 1.05, so aₙ = 1000·(1.05)n−1: 1000, 1050, 1102.50, 1157.63, …. Compare with simple interest's 1000, 1050, 1100, 1150, …: they agree for one year, then compounding quietly pulls ahead. After 10 years compound interest reaches $1628.89 while simple interest gives only $1500 — the gap is interest that earned interest. (The lists above count the $1000 principal as the n = 0 term; the figures below instead plot the end-of-year balance, so their first dot is at n = 1 — the year-1 value $1050 — not the principal.)
Growth: q > 1 (compounding, breeding populations). Decay: 0 < q < 1 (half-life, depreciation). Radioactive half-life means q = ½ exactly: 80 g → 40 → 20 → 10 → 5, halving every period. The rule is always the same — multiply, don't add.
Dial the yearly rate and the number of years. Amber stems are simple interest (arithmetic, collinear); green stems are compound interest (geometric, curving up). Both start from $1000.
A loan is the most natural recurrence in everyday life: this month's balance is born from last month's. Borrow b₁ dollars at monthly rate r; each month the lender first adds interest (multiply by 1 + r), then you make a fixed payment (subtract it). So
bₙ = bₙ₋₁(1 + r) − payment, n ≥ 2.
This names how to get each balance from the one before — it does not name the value directly. To find a closed general term, peel the recurrence around its fixed point b* = payment / r (the balance that would stay put because its interest exactly equals the payment). Writing cₙ = bₙ − b* turns the recurrence into a clean geometric one, cₙ = cₙ₋₁(1 + r), so cₙ = c₁(1 + r)n−1 and
bₙ = b* + (b₁ − b*)(1 + r)n−1, b* = paymentr.
Borrow b₁ = $10000 at r = 1% a month, paying $500 a month. First the recurrence, term by term: b₂ = 10000·1.01 − 500 = 10100 − 500 = $9600; b₃ = 9600·1.01 − 500 = $9196; b₄ = $8787.96. The fixed point is b* = 500 / 0.01 = $50000. Here the gap (b₁ − b*) = 10000 − 50000 = −40000 is negative, and the factor (1.01)n−1 > 1 magnifies that gap (it runs −40000, −40400, −40804, …). Magnifying a negative gap pulls the balance steadily down below b*, so it must hit 0 — the loan gets paid off. Pay less than the interest $100/month and b* would sit below b₁, making the gap positive; magnifying a positive gap then sends the balance up forever.
Borrow $10,000 at 1% a month. Dial the monthly payment and read how many months it takes — every balance is built by the recurrence bₙ = bₙ₋₁(1.01) − payment, never guessed.
If your payment is smaller than the very first month's interest (here $100), the balance rises every month and never closes — the gap to the fixed point grows by (1 + r)n−1. A recurrence can blow up just as easily as it can settle; reading the fixed point tells you which.
Not every sequence is arithmetic or geometric. Some are pure recurrence. The most famous, posed by Leonardo of Pisa (Fibonacci) about breeding rabbits, adds the two previous terms:
a₁ = 1, a₂ = 1, aₙ = aₙ₋₁ + aₙ₋₂ → 1, 1, 2, 3, 5, 8, 13, 21, 34, …
It appears in petals, pinecones, sunflower spirals, and pineapple skins. But the magic is in the ratio of neighbours aₙ₊₁ / aₙ. Compute them: 1, 2, 1.5, 1.667, 1.6, 1.625, 1.615, 1.619, … They bounce above and below, but the swings shrink and the ratios creep toward a single number, the golden ratio
φ = 1 + √52 ≈ 1.618.
Why φ? If the ratios settle at some value L, then dividing aₙ₊₁ = aₙ + aₙ₋₁ by aₙ gives aₙ₊₁/aₙ = 1 + aₙ₋₁/aₙ, i.e. L = 1 + 1/L, so L² = L + 1. The positive root is exactly φ. That word "settle at" is a limit — the very first one in this course.
Show more terms and watch the ratio dots aₙ₊₁/aₙ zigzag tighter and tighter onto the green φ ≈ 1.618 line — a sequence approaching a limit.
The first ratios are 1, 2, 1.5 — wildly different. If you stopped at three you'd guess anything. A limit is about the tail of a sequence, not its opening: never read a destination off the first handful of terms.
Look back at what a sequence is. A term aₙ carries a size (its value) and an order (its index n). That is already more than a lone number. The next strand, Stage 27 · Plane Vectors, gives each number a third thing — a direction — by pairing it with a partner: an ordered pair (x, y) that you can also picture as an arrow. A sequence of numbers becomes a sequence of arrows, and addition stops meaning "combine totals" and starts meaning "follow one arrow, then the next."
And the second gift is the word that ran through the whole stage. The geometric running total a₁/(1 − q) that the partial sums approach; the Fibonacci ratio that approaches φ — both are sequences creeping toward a value they never quite reach. Make that idea precise and you have invented the limit, and with it the slope of a curve and the area under it: calculus. So the humble row of dots you started with in 26.1 is, all at once, a savings plan, a loan, a spiral of rabbits — and the doorway to the two biggest ideas in the rest of mathematics.
Read the shape of the process first: does each period add the same amount, or multiply by the same factor, or is each value born from the previous ones? That single question chooses your model.
| Real-world process | Model | Each step | Key formula |
|---|---|---|---|
| Save $50/mo · flat raise · simple interest | arithmetic | + d | aₙ = a₁ + (n−1)d, Sₙ = n(a₁+aₙ)/2 |
| Compound interest · population · half-life | geometric | × q | A = P(1+r)n, aₙ = a₁qn−1 |
| Loan / installment balance | recurrence | ×(1+r) − pay | bₙ = bₙ₋₁(1+r) − pay |
| Fibonacci (rabbits, spirals) | recurrence | add two prior | aₙ = aₙ₋₁ + aₙ₋₂, aₙ₊₁/aₙ → φ ≈ 1.618 |
| Confusing the two interest types | trap | add ≠ × | simple is a line; compound curves |
Every model's domain is the whole periods n ∈ ℕ — plot dots, not a line. And the word that closes the stage, limit, opens the next: a number about to gain a direction (a vector), and "creeping toward a value" about to become calculus.
You save $80 at the end of every month. Is your balance arithmetic or geometric, and what is it after 10 months?
Arithmetic — the same $80 is added each month, a constant difference d = 80. The balance is the running total Sₙ. With a₁ = 80 and a₁₀ = 80·10 = 800, S₁₀ = 10·(80 + 800)/2 = 10·440 = $4400.
$2000 is invested at 6% compounded yearly. Write the value after n years, and find the value after 3 years.
Geometric with ratio q = 1.06: A = 2000·(1.06)n. After 3 years, A = 2000·(1.06)³ = 2000·1.191016 = $2382.03. (Simple interest would give only 2000 + 2000·0.06·3 = $2360 — compounding adds the extra $22.03.)
A 64 mg sample of an isotope has a half-life of 1 year. Write the recurrence and the general term for the mass after n years, and find the mass after 4 years.
Geometric with q = ½. Recurrence: m₁ = 64, mₙ = mₙ₋₁·½. General term: mₙ = 64·(½)n−1, so the mass after n years is 64·(½)n. After 4 years: 64·(½)⁴ = 64/16 = 4 mg (64 → 32 → 16 → 8 → 4).
A loan of $5000 charges 2% a month and you pay $300 a month. Write the recurrence, compute b₂ and b₃, and decide whether the loan will ever be paid off.
Recurrence: bₙ = bₙ₋₁·1.02 − 300. b₂ = 5000·1.02 − 300 = 5100 − 300 = $4800; b₃ = 4800·1.02 − 300 = 4896 − 300 = $4596. Fixed point b* = 300/0.02 = $15000. Since b₁ = $5000 < $15000, the balance shrinks each month — yes, it gets paid off. (The first month's interest is only $100, well under the $300 payment.)
Using Fibonacci 1, 1, 2, 3, 5, 8, 13, 21, …, compute the ratios a₇/a₆ and a₈/a₇. What single number are they approaching, and why is it not exactly reached at any finite step?
a₆ = 8, a₇ = 13, a₈ = 21. So a₇/a₆ = 13/8 = 1.625 and a₈/a₇ = 21/13 ≈ 1.6154. They straddle and approach φ = (1+√5)/2 ≈ 1.618. Every Fibonacci ratio is a ratio of whole numbers (rational), but φ is irrational, so no finite ratio can equal it — the sequence only converges to it as a limit.
After 12 years, which is larger: $1000 earning 8% simple interest, or $1000 earning 8% compound yearly? Estimate by how much.
Simple: 1000 + 1000·0.08·12 = 1000 + 960 = $1960. Compound: 1000·(1.08)12 = 1000·2.51817 ≈ $2518.17. Compound is larger by about $558. The longer the time, the wider the gap, because compounding earns interest on its own interest — multiplying always overtakes adding.
Six questions to lock it in. Tap the answer you think is right.
This closing lesson applies arithmetic and geometric sequences to finance and natural growth, addressing HSF-BF.A.1 and HSF-BF.A.2 (build a function — explicit or recursive — that models a relationship, and translate between the two forms) and HSF-LE.A.1–A.2 (distinguish situations that change by equal differences vs. equal factors, and construct linear and exponential models). Compound interest A = P(1+r)n and the loan recurrence bₙ = bₙ₋₁(1+r) − pay also touch HSA-SSE.B.3–B.4 (interpreting and using closed-form expressions for series). The Fibonacci-ratio limit toward φ informally previews the limit concept underlying calculus, and rewards MP.7–MP.8 (looking for and expressing regularity in repeated reasoning).