If $x,y,z$ are three natural numbers are in A.P. and $x+y+z=21$ then the possible number of values of the ordered triplet $(x,y,z)$ is I have assumed $x=a-d,y=a$ and $z=a+d$ by which I get $a=7$  and numbers could be $1,7,13$ or $13,7,1$ or $7,7,7$,  but the total no of solutions set is $13$.
I am not getting it. Please help.

Since $z-y=y-x$. $x+z=2y$ and hence $3y=21$. So, $y=7$.
$x+z=2y=14$. So, $z=14-x$.
For any value of $x$, $x,7,14-x$ are in A.P.
If a natural number means a positive integer, them $x$ can take the values $1,2,3,\dots,13$.
