Creating sequences from natural numbers How much different arithmetic sequences can you make from the numbers 1 to 51 ?
Note:
The sequence length has to be 3 numbers.
The difference between each 2 numbers is positive.
 A: There are several ways to count in an organized manner.. We describe three of them.
(Our favourite way is the first, and relatives.)
First way: We need to choose two numbers $a$ and $b$ to serve as the ends of our sequence. The numbers $a$ and $b$ determine a three-term increasing arithmetic sequence  precisely if $\frac{a+b}{2}$ is an integer. So we want $a$ and $b$ to have the same parity.
There are $25$ even numbers in our interval. We can choose $2$ of them in $\binom{25}{2}$ ways. There are $26$ odd numbers in the interval. We can choose $2$ of them in $\binom{26}{2}$ ways. 
Thus the total is $\binom{25}{2}+\binom{26}{2}$.
Second way: There is one three-term increasing arithmetic sequence with "middle" number $2$, there are $2$ with middle number $3$, there are $3$ with middle number $4$,  and so on up to $24$ with middle number $25$.
There are $25$ with middle number $26$.
And by symmetry there are just as many with middle number $\gt 26$ as there are with middle number $\lt 26$.
Thus our count is $2(1/2)(24)(25)+25$, that is, $25^2$. 
Third way: There is $1$ (three-term increasing) arithmetic sequence with common difference $25$, there are $3$ with common difference $24$, there are $5$ with common difference $23$, and so on up to $49$ with common difference $1$.
The sum $1+3+5+\cdots+49$ of the odd numbers up to $49$, is, by a standard formula, equal to $25^2$.
A: A valid arithmetic sequence of length $3$ and of constant (integer) difference $d>0$ consists of the numbers
$$
n,n+d,n+2d
$$
For some integer $n$.  We therefore can state that we will have a valid sequence as long as, given $n$ from $1$ to $51$, $n+2d\leq 51$. 
That is, for any $n$, we have a valid sequence if we choose a $d$ such that $2d\leq 51-n$.
For $51,50$, there is no valid choice of $d$.  For $49,48$, there is one valid choice of $d$. Keep going down in this fashion and you find that the pattern continues so that with $n=3,2$, there are $24$ valid choices, and for $n=1$, there are $25$.
Add all these up to get the total number and you have $N$, the number of valid sequences, is
$$
\begin{align}
N &= 1+1+2+2+3+3+\cdots+24+24+25 \\
&= 2\times(1+2+\cdots+24)+25\\
&= 2\frac{24\times25}{2}+25\\
&= 25\times25\\
&= 625
\end{align}
$$
