Mean Sequence
Weekly Problem 38 - 2009
This sequence is given by the mean of the previous two terms. What is the fifth term in the sequence?
This sequence is given by the mean of the previous two terms. What is the fifth term in the sequence?
Age
11 to 14
Challenge level
The first two terms of a sequence are $\frac{2}{3}$ and $\frac{4}{5}$.
Each subsequent term is the mean of the two previous terms.
What is the fifth term in the sequence?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
Answer: $\frac34$
Numbers
$\frac{2}{3}+\frac{4}{5}=\frac{10}{15}+\frac{12}{15}$, so the average is $\frac{11}{15}$
$\frac45+\frac{11}{15} = \frac{12}{15}+\frac{11}{15}=\frac{23}{15}$ so the average is $\frac{23}{30}$
$\frac{11}{15}+\frac{23}{30} = \frac{22}{30}+\frac{23}{30}=\frac{45}{30}$ so the average is $\frac{45}{60}=\frac34$
Algebra
Suppose the first two terms of the sequence are $x$ and $y$.
Third term is $\frac{1}{2}(x+y)$
Fourth term is $\frac12\left(y+\frac12(x+y)\right)=\frac{1}{4}(x+3y)$
Fifth term is $\frac12\left(\frac12(x+y)+\frac14(x+3y)\right)=\frac{1}{8}(3x+5y)$.
Putting $x=\frac{2}{3}$ and $y=\frac{4}{5}$ we obtain $\frac{1}{8}(2+4)=\frac{3}{4}$.