You may also like

problem icon


Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?

problem icon

Happy Numbers

Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.

problem icon

Intersecting Circles

Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?

Mean Sequence

Stage: 3 Short Challenge Level: Challenge Level:1
$\frac{2}{3}$ 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 $\frac{1}{8}(2+4)=\frac{3}{4}$.

This problem is taken from the UKMT Mathematical Challenges.