Suppose the first two terms of the sequence are $x$ and $y$. The third term is $\frac{1}{2}(x+y)$, and the fourth term is $\frac{1}{4}(x+3y)$ so the fifth term is $\frac{1}{8}(3x+5y)$.

Putting $x=\frac{2}{3}$ and $y=\frac{4}{5}$ we obtain $\frac{1}{8}(2+4)=\frac{3}{4}$.

This problem is taken from the UKMT Mathematical Challenges.
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