$\frac{2}{3}$ and $\frac{4}{5}$ can be written as $\frac{10}{15}$ and $\frac{12}{15}$, so their average is $\frac{11}{15}$.

The average of $\frac{12}{15}$ and $\frac{11}{15}$ is $\frac{23}{30}$.

The average of $\frac{11}{15}$ and $\frac{23}{30}$ is $\frac{45}{60} = \frac34$.

Alternatively, 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 $\frac{1}{8}(2+4)=\frac{3}{4}$.

The average of $\frac{12}{15}$ and $\frac{11}{15}$ is $\frac{23}{30}$.

The average of $\frac{11}{15}$ and $\frac{23}{30}$ is $\frac{45}{60} = \frac34$.

Alternatively, 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.*