1. ordered pairs question

if a b c are three natural numbers in Arithmetic progression and a+b+c=21 then all the possible values of the ordered triplet (a,b,c) is

2. If they are in arithmetric progression, a, b and c have a common difference, or:

$c - b = b - a = \dfrac{c - a}{2}$

Then;

$a + b + c = 21$

I think that should be okay to solve now.

3. Originally Posted by prasum
if a b c are three natural numbers in Arithmetic progression and a+b+c=21 then all the possible values of the ordered triplet (a,b,c) is
Your AP is $n,~n+d,~n+2d$ so $n+n+d+n+2d=3n+3d=21$.
That means that $n+d=7$.
The numbers of pairs that satisfy that depends on if we consider 0 a natural or not.

Here is an example: $n=3~\&~d=4$ then the AP is $3,~7,~11$.

Can you finish?