Drill 1 · Math · arithmetic-geometric-sequences
AP Precalculus: Arithmetic & Geometric Sequences (Drill 1) is a Math practice drill covering arithmetic-geometric-sequences. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice arithmetic and geometric sequences for the AP® Precalculus exam with these five questions. Topics include identifying sequence type from tables, writing explicit and recursive formulas, finding missing terms using the geometric mean, and connecting geometric sequences to exponential models.
Question 1. The table below shows selected terms of a sequence.
| n | a(n) |
|---|---|
| 1 | 5 |
| 2 | 20 |
| 3 | 80 |
| 4 | 320 |
Explanation: Choice D is correct. Check successive ratios: 20/5 = 4, 80/20 = 4, 320/80 = 4. Since the ratio between consecutive terms is constant at 4, this is a geometric sequence with common ratio 4. Choice A is incorrect because 15 is the difference between the first two terms (20 − 5 = 15), but the differences are not constant (80 − 20 = 60, 320 − 80 = 240), so the sequence is not arithmetic; the student used the first difference without checking the rest. Choice B is incorrect because 4 is the common ratio, not a common difference; the differences between consecutive terms (15, 60, 240) are not constant, so the sequence is not arithmetic. Choice C is incorrect because the common ratio is 4, not 15; the student identified the correct sequence type (geometric) but used the first difference (20 − 5 = 15) as the ratio instead of the quotient (20 ÷ 5 = 4).
Question 2. An arithmetic sequence has first term a(1) = 7 and common difference d = −3. Which of the following gives the explicit formula for the nth term?
Explanation: Choice D is correct. The explicit formula for an arithmetic sequence is a(n) = a(1) + d(n − 1). Substituting a(1) = 7 and d = −3: a(n) = 7 + (−3)(n − 1) = 7 − 3(n − 1). Verify: a(1) = 7 − 3(0) = 7 ✓, a(2) = 7 − 3(1) = 4 ✓, a(3) = 7 − 3(2) = 1 ✓. Choice A is incorrect because a(n) = 7 − 3n gives a(1) = 7 − 3 = 4, not 7; the student forgot to subtract 1 from n, effectively starting the index at n = 0 rather than n = 1. Choice B is incorrect because a(n) = 7 − 3(n + 1) gives a(1) = 7 − 3(2) = 1, not 7; the student added 1 inside the parentheses rather than subtracting it, shifting the sequence in the wrong direction. Choice C is incorrect because 7(−3)n−1 is the explicit formula for a geometric sequence with first term 7 and common ratio −3; the student applied the geometric formula to an arithmetic sequence, confusing multiplication in the recursive structure with a common ratio.
Question 3. A colony of bacteria is growing geometrically. On day 4, the population is 4,000. On day 8, the population is 49,000. Assuming the population forms a geometric sequence, what is the population on day 6?
Explanation: Choice B is correct. In a geometric sequence, the middle term between two terms equidistant from it is the geometric mean of those two terms. Since day 6 is exactly halfway between day 4 and day 8, the population on day 6 equals √(4,000 × 49,000) = √(196,000,000) = 14,000. Choice A is incorrect because 11,250 = (4,000 + 49,000)/2 ÷ 4 or similar arithmetic reasoning; the student attempted an arithmetic approach rather than using the geometric mean. Choice C is incorrect because 26,500 = (4,000 + 49,000)/2, the arithmetic mean of the two given values; the student averaged the populations as if the sequence were arithmetic rather than geometric. Choice D is incorrect because 53,000 ≈ 49,000 + 4,000, suggesting the student added the two known values; this has no basis in sequence reasoning and confuses the geometric mean with the sum.
Question 4. A geometric sequence has first term a(1) = 12 and common ratio r = \(\frac{1}{3}\). Which of the following exponential functions matches the sequence values for integer inputs n?
Explanation: Choice B is correct. The explicit formula for the sequence is a(n) = 12 · (1/3)n−1. To write this as an exponential function f(x) = a · bx, rewrite: 12 · (1/3)n−1 = 12 · (1/3)n · (1/3)−1 = 12 · 3 · (1/3)n = 36 · (1/3)n. Check: f(1) = 36 · (1/3) = 12 ✓, f(2) = 36 · (1/9) = 4 ✓ (which equals a(2) = 12 · 1/3 = 4). Choice A is incorrect because f(x) = 12 · 3x is an increasing exponential; the sequence is decreasing since r = 1/3 < 1, so the base must be 1/3, not 3. Choice C is incorrect because f(1) = 12 · (1/3) = 4 ≠ 12; this formula gives a(n) = 12 · (1/3)n which does not equal a(1) = 12. The student used the formula a · rn rather than the correct a · rn−1 and failed to adjust the initial coefficient. Choice D is incorrect because f(1) = 4 · (1/3) = 4/3 ≠ 12; the student divided 12 by 3 rather than multiplying by 3 when adjusting the coefficient.
Question 5. A sequence is defined recursively by a(1) = 2 and a(n) = 3 · a(n − 1) for n ≥ 2. Which of the following statements about this sequence is correct?
Explanation: Choice D is correct. Since each term is multiplied by 3 to get the next, this is a geometric sequence with common ratio 3. The first few terms are: a(1) = 2, a(2) = 6, a(3) = 18, a(4) = 54. The explicit formula is a(n) = a(1) · rn−1 = 2 · 3n−1. Verify: a(1) = 2 · 30 = 2 ✓, a(2) = 2 · 31 = 6 ✓, a(3) = 2 · 32 = 18 ✓. Choice A is incorrect because the recursive rule a(n) = 3 · a(n−1) multiplies by 3, not adds 3; the sequence is geometric, not arithmetic. The student confused multiplication in the recursive rule with addition. Choice B is incorrect because a(n) = 2 · 3n gives a(1) = 2 · 3 = 6 ≠ 2; the student used n instead of n−1 as the exponent, shifting the sequence by one term. Choice C is incorrect because a(n) = 6 · 3n−1 gives a(1) = 6 · 1 = 6 ≠ 2; the student used a(2) = 6 as the starting value of the explicit formula rather than a(1) = 2.