Number of three-term arithmetic progressions in [n] 
Three numbers are chosen at random between 1 and $n$ (say $n=500$).What will be the probability of those numbers to be in arithmetic progression?

I don't know how to count the number of favorable events.
Sample space={(1,2,3),(4,5,6),(18,20,21)........} 
favorable events={(2,4,6),(8,12,16),(10,20,30),(50,100,150... and many more)}
I can count the sample space but how do i count the favorable events.
Some insight could help?
 A: (Assuming three distinct numbers are chosen.)
A three-element arithmetic progress $a,b=a+k,c=a+2k$ with $k>0$ is entirely determined by $a,c$, and thus the number of them is equal to the number of ways of picking $a,c$ with the same parity (either both odd or both even.) 
The number of odd numbers from $1$ to $n$ is $\lceil n/2\rceil$ and the number and the number of even numbers from $1$ to $n$ is $\lfloor n/2\rfloor$.
This means that the number of arithmetic progressions is:
$$\binom{\lfloor n/2\rfloor}{2}+\binom{\lceil n/2\rceil}{2}$$
When $n$ is even, this is $2\binom{n/2}{2}=\frac{n(n-2)}{4}$. When $n$ is odd, this is $\binom{(n-1)/2}{2}+\binom{(n+1)/2}2=\frac{(n-1)^2}{4}.$
Both cases can be written in one formula:
$$\left\lfloor\frac{(n-1)^2}{4}\right\rfloor$$
When $n>2$ is even, the probability is:
$$\dfrac{\frac{n(n-2)}{4}}{\frac{n(n-1)(n-2)}{6}}=\frac{3}{2(n-1)}$$
When $n>1$ is odd:
$$\frac{3(n-1)}{2n(n-2)}$$
A: I have assumed that three randomly chosen numbers are distinct.
Let $d$ be the common difference of the arithmetic progression. Then it is easy to see that $d$ must take an integer value between $1$ and $249$ inclusive. So for any one of these $d$ values, let $a$ be the smallest of the trio in the arithmetic progression. Then we just have to count the number of distinct $a$s we can have for each $d$.
First let's consider $d=1$, then $a$ can be between $1$ and $498$. 
For $d=2$, $a$ is between $1$ and $496$.
It quickly becomes evident that the range of values $a$ can take for any given $d$ is from the set $\{1, 2, \cdots , 500-2d\}$, and this is easily shown to be true since if $a$ is the smallest, then $a+2d$ is the largest of the trio and must not exceed $500$
So the number of different trio of APs we can form is: $$\sum_{i=1}^{249} (500-2i)$$ Can you determine this value? (Note that if three randomly chosen numbers can be the same, then the sum takes $i$ from $0$ not $1$, otherwise it should be the same).
Edit: I realized that $n=500$ is an example, and the asker wanted general $n$. In this case $d$ can take values between $1$ (or $0$ if non distinct integers are picked) and $\lfloor{\frac{n-1}{2}}\rfloor$.
In this case we get $$\sum_{i=1}^{\lfloor{\frac{n-1}{2}}\rfloor}n-2i$$
A: Counting favorable samples:
$(1,1+k,1+2k)$ is favorable.  So are $(2,2+k,2+2k),...,(n-2k,n-k,n)$.  For an increment of $k$, there are $n-2k$ such combinations.  The increment can take all values from 1 to $[\frac{n-1}{2}]$ where the square brackets indicate "floor", i.e., $\frac{n-1}{2}$ if n is odd and $\frac{n-2}{2}$ if $n$ is even.   Total number of favorable combinations is $$\sum_{k=1}^{[\frac{n-1}{2}]}n-2k$$ which is easily summed as $\frac{n^2-2n+1}{4}$ if $n$ is odd and $\frac{n^2-2n}{4}$ if $n$ is even.
A: Here's another approach. 
The number of ways of selecting $3$ numbers from $n$ is $N=\binom n3=\frac 16 n(n-1)(n-2)$.
For any $3$ numbers to be in AP, they must be successively equidistant.
Let $f(s)$ be the number of equidistant triplets including $s$ itself, which can be formed from the first $s$ positive integers. 
Counting backwards from $s$, it can be easily established that
$$f(2r-1)=f(2r)=r-1$$
The total number of equidistant triplets which can be formed from the first $n$ integers (both including and excluding $n$ itself) is 
$$S_n=\sum_{r=3}^n f(r)$$
For even $n$,
e.g. $n=2m$ (i.e.$m=\frac n2$), 
$$\begin{align}
S&=\sum_{r=3}^{2m}f(r)\\
&=\sum_{r=2}^m\underbrace{f(2r-1)}_{r-1}+\underbrace{f(2r)}_{r-1}\\
&=2\sum_{r=2}^m(r-1)\color{lightgrey}{=2\sum_{r=1}^{m-1}r=2\binom m2}\\
&=m(m-1)\color{lightgrey}{=\frac n2\left(\frac n2-1\right)}\\
&=\frac 14{n(n-2)}\end{align}$$
Probability of selecting $3$ numbers which are in AP is
$$\frac SN=\frac{\frac 14n(n-2)}{\frac 16n(n-1)(n-2)}=\color{red}{\frac 3{2(n-1)}}$$
For odd $n$,
  e.g. $n=2m-1$ (i.e.$m=\frac {n+1}2$), 
$$\begin{align}
S&=\sum_{r=3}^{2m-1}f(r)\\
&=\left(\sum_{r=3}^{2m}f(r)\right)-\underbrace{f(2m)}_{m-1}\\
&=\frac 14(n+1)(n-1)-\left(\frac {n+1}2-1\right)\\
&=\frac 14{(n-1)^2}
\end{align}$$
Probability of selecting $3$ numbers which are in AP is
$$\frac SN=\frac{\frac 14(n-1)^2}{\frac 16n(n-1)(n-2)}=\color{red}{\frac 3{2n(n-2)}}$$
