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TMUA 2022 · Paper 1 · Question 5 of 20

TMUA 2022 Paper 1 Question 5

Sequences and series — Recurrence relations · finding the parameters. Try it first; the answer and a full worked solution are below.

TMUA 2022 · Paper 1Sequences and seriesRecurrence relations · finding the parameters8 options
The terms xn of a sequence follow the rule xn+1 = xn + pxn + q where p and q are real numbers.

Given that x1 = 3, x2 = 5 and x3 = 7, find the value of x4.

  1. A−5
  2. B5
  3. C517
  4. D152
  5. E233
  6. F9
  7. G11
  8. H13
Show the answer and worked solution
answer · H
  1. A−5
  2. B5
  3. C517
  4. D152
  5. E233
  6. F9
  7. G11
  8. H13
From x2 = 5: 3+p3+q = 5, so p = 12 + 5q. From x3 = 7: 5+p5+q = 7, so p = 30 + 7q. Equating, 12 + 5q = 30 + 7q, giving q = −9 and then p = −33. So x4 = 7 − 337 − 9 = −26−2 = 13.