IMC 2008 Q16
16. The first two terms of a sequence are \(\dfrac{2}{3}\) and \(\dfrac{4}{5}\). Each term after the second term is the average (mean) of the two previous terms. What is the fifth term in the sequence?
- A \(\dfrac{5}{34}\)
- B \(\dfrac{1}{2}\)
- C \(\dfrac{10}{13}\)
- D \(\dfrac{3}{4}\)
- E \(\dfrac{10}{11}\)
Intermediate Mathematical Challenge 2008, Question 16. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D \(\dfrac{3}{4}\)
The problem may be solved by firstly calculating the third and fourth terms of the sequence, but an algebraic method does reduce the amount of calculation involved. Let the first two terms be \(x\) and \(y\) respectively. Then the third term is \(\tfrac{1}{2}(x + y)\), whilst the fourth term is \(\tfrac{1}{4}(x + 3y)\). So the fifth term is \(\tfrac{1}{8}(3x + 5y)\). Putting \(x = \dfrac{2}{3}\) and \(y = \dfrac{4}{5}\), we obtain \(\dfrac{2 + 4}{8} = \dfrac{3}{4}\).
Solution © UK Mathematics Trust, from the IMC 2008 solutions.