aₙ = aₖ + d(n − k) and gₙ = gₖ · r^(n−k) from any given term — not just the firstTwo job offers, both starting at $50,000:
Year 1: A pays $53,000, B pays $52,000 — A is ahead. Year 10: A pays $80,000; B pays 50,000(1.04)¹⁰ ≈ $74,012 — still A. Year 20: A: $110,000; B: 50,000(1.04)²⁰ ≈ $109,556. Year 21: B takes the lead, and it never looks back.
Adding beats multiplying early; multiplying wins forever after. That crossover — and the formulas that locate it — is Unit 2's opening act.
An arithmetic sequence changes by a constant common difference d each step:
aₙ = a₀ + d·n (indexed from a₀)
aₙ = aₖ + d·(n − k) (from any known term aₖ)
The second form is the workhorse: it says "start where you know, add d for each step you travel." Given two terms, d = (change in value)/(change in index) — a slope computation. Arithmetic sequences ARE linear functions sampled at integers: same constant rate of change, same graph pattern (collinear points).
A geometric sequence changes by a constant common ratio r each step:
gₙ = g₀ · rⁿ (indexed from g₀)
gₙ = gₖ · r^(n − k) (from any known term gₖ)
Given two terms, r^(steps between them) = value ratio, so r is a root: if g₂ = 18 and g₅ = 486, then r³ = 486/18 = 27 → r = 3. Geometric sequences are exponential functions sampled at integers.
Percent language converts to r: - "grows 4% per year" → r = 1.04 - "loses 12% per year" → r = 0.88 (keep 88%) - 0 < r < 1: decay; r > 1: growth; r = 1: constant
Over equal steps:
| Check | Constant? | Family |
|---|---|---|
| differences aₙ₊₁ − aₙ | yes | arithmetic / linear |
| ratios aₙ₊₁ / aₙ | yes | geometric / exponential |
Sequence 4, 12, 36, 108: differences 8, 24, 72 (no); ratios 3, 3, 3 (yes) → geometric.
An increasing exponential (r > 1) grows by a constant percent of an ever-larger amount; a linear function grows by a constant amount. Each step, the exponential's raise gets bigger while the linear's stays fixed — so eventually the exponential's single-step gain exceeds the linear's, and the gap only widens. Formally: for any linear f and any exponential g with r > 1, g(n) > f(n) for all sufficiently large n. (You watched it happen in the warm-up at year 21.)
[GRAPH: Discrete points for two sequences vs. n: arithmetic 50 + 3n rising as evenly spaced dots along a line; geometric 50(1.04)ⁿ hugging below it at first, crossing near n = 21, then pulling away upward. Crossover point circled and labeled "exponential overtakes — permanently".]
Problems index from a₀ ("initial value") or a₁ ("first term") — read carefully. The safe universal habit: use the known-term form aₖ + d(n − k) or gₖ·r^(n−k), which never cares where counting started.
Problem: (a) Arithmetic with a₁ = 7, d = 4: find a₁₀. (b) Geometric with g₀ = 5, r = 2: find g₆.
Solution: (a) a₁₀ = a₁ + d(10 − 1) = 7 + 4(9) = 43. (b) g₆ = 5·2⁶ = 5·64 = 320.
Interpretation: Count steps, not terms: from index 1 to index 10 is 9 steps. The off-by-one error here is the single most common sequence mistake.
Problem: An arithmetic sequence has a₃ = 11 and a₇ = 27. Find a₁₀.
Solution: d = (27 − 11)/(7 − 3) = 16/4 = 4. From the known term a₇:
a₁₀ = a₇ + d(10 − 7) = 27 + 4(3) = 39
Interpretation: Slope between known points, then travel from the nearer one. You never need a₀ or a₁.
Problem: A geometric sequence has g₂ = 18 and g₅ = 486. Find r and g₀, and write the nth-term formula.
Solution: Three steps from index 2 to 5: r³ = 486/18 = 27 → r = 3. Walk backwards: g₀ = g₂/r² = 18/9 = 2. Formula: gₙ = 2·3ⁿ. Check g₅: 2·243 = 486 ✓
Interpretation: Ratios across multiple steps compound: k steps → rᵏ. Take the root, then anchor the formula at any known term.
Problem: Town A's population is 40,000 and grows by 1,200 people per year. Town B's is 25,000 and grows by 5% per year. Model each town's population n years from now and find when B first exceeds A.
Solution: A(n) = 40,000 + 1,200n (arithmetic/linear); B(n) = 25,000(1.05)ⁿ (geometric/exponential). Solve B(n) > A(n) numerically: graph both, find intersection 📱 → n ≈ 19.98... check n = 19: A = 62,800; B = 25,000(1.05)¹⁹ ≈ 63,163 — B already ahead at 19. Check n = 18: A = 61,600; B ≈ 60,155 — A still ahead. So B first exceeds A in year 19.
Interpretation: For "first year that…" questions with integer years, the graphing intersection guides you, but verify the neighboring integers — the crossing rarely lands on a whole number, and the exam wants the first integer past it.
1. (B). a₁₀ = 7 + 4(9) = 43. (A) uses 10 steps (off-by-one); (D) uses 4 + 7·4 backwards.
2. (A). 5·2⁶ = 320. (B) uses 2⁵; (C) 2⁷; (D) computes 5·2·6.
3. (D). d = 5, a₁ = 3: aₙ = 3 + 5(n − 1) = 5n − 2. Check n = 1: 3 ✓. (B) fails at n = 1 (gives 8).
4. (C). 48/96 = 1/2. (B) inverts; negatives would need alternating signs.
5. (A). d = 4; a₁₀ = a₇ + 3d = 27 + 12 = 39 (Example 2). (B) takes 4 steps; (C) takes 2.
6. (B). r³ = 27 → r = 3; g₀ = g₂/r² = 18/9 = 2 (Example 3). (A) divides by r once; (C) confuses g₀ with r.
7. (C). Ratios 12/4 = 36/12 = 108/36 = 3. (A) checks only the first difference; (B) would need constant second differences (they're 16, 48 — no).
8. (B). Exponential with r > 1 eventually dominates any linear — each 10% raise grows while +5000 stays fixed. (A) confuses "larger now" with "larger always."
9. (D). Keeps 88% each year: geometric, r = 0.88. (B) is the keep-vs-lose reversal; (C) treats the first year's dollar loss as if constant (12% of a shrinking value shrinks too).
10. (C). "3 more than the previous" = constant difference 3 → arithmetic. (A) mistakes "more" for "times."
11. (D). r = 1 + 0.08 = 1.08. (A) forgets the 1; (C) is an 8% decrease.
12. (FRQ-style, 6 points) (i) [2 pts] Plan L: Pₙ = 400 + 60n. Plan G: Pₙ = 400(1.1)ⁿ. (ii) [2 pts] L: 400 + 300 = 700 fish. G: 400(1.61051) = 644.204 → about 644 fish. (iii) [2 pts] Plan G. Plan L adds a constant 60 fish/year, while Plan G multiplies by 1.1 — its yearly gain (10% of an ever-growing population) increases every year, so it eventually exceeds 60 and keeps growing; from that year on G's total gain outstrips L's, and an increasing geometric sequence exceeds any arithmetic sequence for all sufficiently large n.
12. (FRQ-style) 🚫 A pond is stocked with 400 fish. Plan L adds 60 fish each year. Plan G increases the population by 10% each year. (i) Write the nth-term formula for the population under each plan (year 0 = 400 fish). (ii) Compute both populations at year 5. (1.1⁵ = 1.61051) (iii) Which plan produces the larger population in the long run? Justify by comparing the growth patterns, not by computing more values.
An arithmetic sequence aₙ and a geometric sequence gₙ satisfy: a₂ = 14, a₆ = 26, g₁ = 6, g₃ = 54, with all terms of gₙ positive.
(a) (i) Find the common difference of aₙ and write a formula for aₙ using the known-term form. (ii) Find a₁₂.
(b) (i) Find the common ratio of gₙ and write a formula for gₙ. (ii) Find g₆.
(c) A student computes a₁ = 11 and g₁ = 6 and claims: "since a₁ > g₁, the arithmetic sequence stays larger for all n." (i) Evaluate the claim by comparing both sequences at n = 3. (ii) State the general principle about arithmetic vs. geometric growth this illustrates.
(a) [2 pts] (i) [1 pt] d = (26 − 14)/(6 − 2) = 12/4 = 3; aₙ = a₂ + 3(n − 2) = 14 + 3(n − 2). (ii) [1 pt] a₁₂ = 14 + 3(10) = 44.
(b) [2 pts] (i) [1 pt] r² = g₃/g₁ = 54/6 = 9 → r = 3 (positive terms → r = 3, not −3); gₙ = 6·3^(n−1). (ii) [1 pt] g₆ = 6·3⁵ = 6·243 = 1458.
(c) [2 pts] (i) [1 pt] At n = 3: a₃ = 14 + 3(1) = 17, while g₃ = 54. The geometric sequence has already overtaken the arithmetic one by the third term, so the claim is false — a head start at n = 1 guarantees nothing. (ii) [1 pt] An increasing geometric sequence (r > 1) eventually exceeds any arithmetic sequence, no matter which starts larger: constant-factor growth outpaces constant-amount growth for all sufficiently large n.
1. (B). a₁₀ = 7 + 4(9) = 43. (A) uses 10 steps (off-by-one); (D) uses 4 + 7·4 backwards.
2. (A). 5·2⁶ = 320. (B) uses 2⁵; (C) 2⁷; (D) computes 5·2·6.
3. (D). d = 5, a₁ = 3: aₙ = 3 + 5(n − 1) = 5n − 2. Check n = 1: 3 ✓. (B) fails at n = 1 (gives 8).
4. (C). 48/96 = 1/2. (B) inverts; negatives would need alternating signs.
5. (A). d = 4; a₁₀ = a₇ + 3d = 27 + 12 = 39 (Example 2). (B) takes 4 steps; (C) takes 2.
6. (B). r³ = 27 → r = 3; g₀ = g₂/r² = 18/9 = 2 (Example 3). (A) divides by r once; (C) confuses g₀ with r.
7. (C). Ratios 12/4 = 36/12 = 108/36 = 3. (A) checks only the first difference; (B) would need constant second differences (they're 16, 48 — no).
8. (B). Exponential with r > 1 eventually dominates any linear — each 10% raise grows while +5000 stays fixed. (A) confuses "larger now" with "larger always."
9. (D). Keeps 88% each year: geometric, r = 0.88. (B) is the keep-vs-lose reversal; (C) treats the first year's dollar loss as if constant (12% of a shrinking value shrinks too).
10. (C). "3 more than the previous" = constant difference 3 → arithmetic. (A) mistakes "more" for "times."
11. (D). r = 1 + 0.08 = 1.08. (A) forgets the 1; (C) is an 8% decrease.
12. (FRQ-style, 6 points) (i) [2 pts] Plan L: Pₙ = 400 + 60n. Plan G: Pₙ = 400(1.1)ⁿ. (ii) [2 pts] L: 400 + 300 = 700 fish. G: 400(1.61051) = 644.204 → about 644 fish. (iii) [2 pts] Plan G. Plan L adds a constant 60 fish/year, while Plan G multiplies by 1.1 — its yearly gain (10% of an ever-growing population) increases every year, so it eventually exceeds 60 and keeps growing; from that year on G's total gain outstrips L's, and an increasing geometric sequence exceeds any arithmetic sequence for all sufficiently large n.
🎯 Exam tip: Translate every sequence word problem into one of two sentences before touching a formula: "same amount each step" (arithmetic — write d with its sign) or "same factor each step" (geometric — convert any percent to r = 1 ± rate). Half the errors in this topic happen before any math does.