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TMUA 2019 · Paper 1 · Question 4 of 20

TMUA 2019 Paper 1 Question 4

Sequences and series — Recurrence with square roots · repeated halving of an index. Try it first; the answer and a full worked solution are below.

TMUA 2019 · Paper 1Sequences and seriesRecurrence with square roots · repeated halving of an index8 options
The sequence xn is given by x1 = 10 xn+1 = √xn   for  n ≥ 1 What is the value of x100?

[Note that abc means a(bc).]

  1. A10299
  2. B102100
  3. C102−99
  4. D102−100
  5. E10−299
  6. F10−2100
  7. G10−2−99
  8. H10−2−100
Show the answer and worked solution
answer · C
  1. A10299
  2. B102100
  3. C102−99
  4. D102−100
  5. E10−299
  6. F10−2100
  7. G10−2−99
  8. H10−2−100
Write every term as a power of 10. Taking a square root halves the exponent, so x1 = 101, x2 = 101/2, x3 = 101/4, and in general xn = 10(1/2)n−1. Hence x100 = 10(1/2)99 = 102−99. Note the terms are all bigger than 1 and shrinking towards 100 = 1, which rules out every negative-base option immediately.