A question featured on a practice test reads:
An infinite increasing list of numbers, has the property that the median of the first $n$ terms equals the $n$th odd positive integer. How many numbers in the list are less than $2018$?
If I start the list with $a$ for example, according to the original question it would be the first odd positive integer, since the median of the 1st term is the same as the 1st term. If I added a second term $b$, then the median (in this case equal to the average) of $a,b$ would be equal to 2nd odd positive integer. How can I apply actual numbers in this list to find how many are less than 2018?
Let's call this special sequence $(a_n)$. We can see immediately that $a_1 = 1$ - the median of a data set of one point is that point, and the first odd number is $1$. Since the median of $a_1,a_2$ is $3$, the second odd number, this necessitates $a_2 = 5$. We can continue extending the sequence like this, using medians further and further down in the sequence to determine the terms. For a few $n$ I determined
$$1,5,9,13,21,25,29...$$
Now, if we assume this pattern to hold (proving it seems a bit beyond me at the moment), we conclude that $a_n = 4(n-1)+1$.
Then the question becomes, for which $n$ is $a_n \ge 2018$? The terms in the sequence all satisfy $a_n \equiv 1 \pmod 4$, so the first term the sequence that exceeds $2018$ will have $a_n = 2021$. Then we see:
$$2021 = 4(n-1) + 1 = 4n - 3 \implies 4n = 2024 \implies n = 506$$
Thus, the $506^{th}$ term is the first to exceed $2018$. Thus, $505$ terms are in this sequence which are less than $2018$.