the number of different triangles How many  different  triangles in which the measurement of the interior angles in degree is representing natural numbers in arithmetic sequence 
 A: I assume you want to know how many triangles up to similarity there are (so side lengths don't matter). I further assume you mean that the interior angles are (in degrees) $n,n+m,n+2m,$ with each a natural number. In particular, we need $n$ to be a natural number, and $m$ to be (without loss of generality) a non-negative integer (unless you require the angles be distinct, in which case we need $m$ positive).
Then we'll need $$180=n+(n+m)+(n+2m)=3n+3m=3(n+m),$$ so $$60=n+m,$$ and so $$m=60-n.$$
Can you get the rest of the way?
A: I believe you meant that if the angle of the triangle are $\left\{\alpha - k, \alpha, \alpha + k\right\}$ which are in arithmetic progression then the sum of these angles $=180^0$. Note, if three numbers are in AP then the the difference between the consecutive terms is constant.
$$\alpha - k + \alpha + \alpha + k = 180$$
So you have to just solve for $\alpha$ in the above equation
On the other hand, if you are interested in the fact that the angles are part of an Arithmetic sequence such that if the angles of a triangle are $\{\alpha,\beta,\gamma\}$ and 
$$\beta = \alpha + nk\,, \gamma = \alpha + mk$$
then
$$\alpha+\beta+\gamma = \alpha+\alpha+nk+\alpha+mk=180^o$$
$$3\alpha+k(n+m)=180^0$$
$$k(n+m)=3(180 - 3\alpha)$$
So if $k=3$ and $1 \le \alpha \le 59$ we will have the set of triangles whose angles are natural numbers and part of an Arithmetic Sequence
A: Let $x,x-a,x+a$ be the tree angles with $a,b \in \Bbb N$. They must verify
$$3x = x + x-a + x+a = 180$$
This yields $x=60$. Now, notice $60-a \geq 1$ iff $a \in \{1,2,\dots,59\}$. Each choice gives exactly one triangle (up to isometry) so the answer is $59$.
Remark. It will be $60$ if you allow the case $a = 0$ (that is $60,60,60$).
