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TMUA 2023 · Paper 1 · Question 18 of 20

TMUA 2023 Paper 1 Question 18

Sequences and series — Geometric series · convergence and probability. Try it first; the answer and a full worked solution are below.

TMUA 2023 · Paper 1Sequences and seriesGeometric series · convergence and probability8 optionshard
You are given that S = 4 + 8k7 + 16k249 + 32k3343 + ⋯ + 4(2k7)n + ⋯ The value of k is chosen as an integer in the range −5 ≤ k ≤ 5. All possible values of k are equally likely to be chosen.

What is the probability that the value of S is a finite number greater than 3?

  1. A111
  2. B110
  3. C311
  4. D310
  5. E511
  6. F12
  7. G711
  8. H710
Show the answer and worked solution
answer · E
  1. A111
  2. B110
  3. C311
  4. D310
  5. E511
  6. F12
  7. G711
  8. H710
The series is geometric with first term 4 and ratio 2k7, so it converges exactly when |2k7| < 1, that is |k| < 3.5, and then S = 41 − 2k7 = 287 − 2k. For the integers −3 ≤ k ≤ 3 the denominator is positive, and S > 3 requires 28 > 3(7−2k), so k > −76. That leaves k ∈ {−1,  0,  1,  2,  3} — five values out of the eleven available, a probability of 511.