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Math 20-1

Math 20-1 · Worksheets

Sequences and series

Ten questions of mixed difficulty, covering Sequences and series. Print it, or work through it on screen — the answer key starts on its own page.

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Sequences and series

Math 20-1 · maddyhelps.com

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Circle the best answer for each question. Show your work in the space provided.

  1. Which sequence is arithmetic?

    1. a) 1, 4, 9, 16
    2. b) 2, 6, 18, 54
    3. c) 1, 1, 2, 3
    4. d) 5, 9, 13, 17

  2. In an arithmetic sequence, t₁ = 4 and t₈ = 32. The common difference is

    1. a) 4
    2. b) 28
    3. c) 3.5
    4. d) 7

  3. The sum of the first 6 terms of 2, 6, 18, … is

    1. a) 728
    2. b) 1458
    3. c) 729
    4. d) 486

  4. The sum of the infinite series 12 + 6 + 3 + 1.5 + … is

    1. a) 24
    2. b) 21
    3. c) ∞
    4. d) 18

  5. The sum of the first 20 terms of 3, 7, 11, … is

    1. a) 820
    2. b) 410
    3. c) 800
    4. d) 1600

  6. If 5, x, 45 are three consecutive terms of a geometric sequence, then x is

    1. a) 15 only
    2. b) 15 or −15
    3. c) 25
    4. d) 9

  7. An arithmetic sequence has t₁ = 5 and d = 3. Then t₁₀ is

    1. a) 35
    2. b) 32
    3. c) 53
    4. d) 30

  8. An infinite geometric series converges when

    1. a) t₁ > 0
    2. b) −1 < r < 1
    3. c) r > 0
    4. d) r > 1

  9. For the sequence 3, 7, 11, 15, …, the common difference is

    1. a) 4
    2. b) −4
    3. c) 7
    4. d) 3

  10. A geometric sequence has t₁ = 2 and r = 3. Then t₅ is

    1. a) 486
    2. b) 162
    3. c) 30
    4. d) 243

Answer key · Sequences and series

Math 20-1 · maddyhelps.com

  1. d) 5, 9, 13, 17 — Arithmetic means a constant difference, and 5, 9, 13, 17 adds 4 each time. The squares 1, 4, 9, 16 have growing differences, 2, 6, 18, 54 is geometric, and 1, 1, 2, 3 is Fibonacci.
  2. a) 4 — From t₈ = t₁ + 7d: 32 = 4 + 7d, so 7d = 28 and d = 4. There are seven gaps between the first and eighth terms, not eight.
  3. a) 728 — Use Sₙ = t₁(rⁿ − 1)/(r − 1) = 2(3⁶ − 1)/2 = 3⁶ − 1 = 728. The sixth term alone is 486, so a total of 728 is the right size.
  4. a) 24 — The ratio is ½, and |r| < 1, so the series converges to S = t₁/(1 − r) = 12/0.5 = 24. Adding forever gives a finite total here because the terms shrink fast enough.
  5. a) 820 — Use Sₙ = n/2[2t₁ + (n − 1)d] = 10[6 + 19(4)] = 10(82) = 820. You can sanity-check it: the average term is about 41, and 20 × 41 = 820.
  6. b) 15 or −15 — In a geometric sequence the ratio between consecutive terms is constant, so x/5 = 45/x and x² = 225. That gives x = 15 (ratio 3) and also x = −15 (ratio −3), and both sequences are genuinely geometric.
  7. b) 32 — Use tₙ = t₁ + (n − 1)d: 5 + 9(3) = 32. The n − 1 matters — you take nine steps to get from the first term to the tenth, not ten.
  8. b) −1 < r < 1 — The terms have to shrink towards zero, which happens exactly when the ratio is between −1 and 1. At r = 1 the terms never shrink and at r = −1 they flip forever, so both endpoints are excluded.
  9. a) 4 — Subtract any term from the one after it: 7 − 3 = 4, and 11 − 7 = 4. The common difference is what you add, so it is a difference, not the first term.
  10. b) 162 — Use tₙ = t₁r^(n−1): 2 × 3⁴ = 2 × 81 = 162. Using 3⁵ gives 486 and is the same off-by-one that catches people in the arithmetic formula.