Is it AP-ish?

We write a 3-digit number as $abc$ .

Having one digit being the average of the two others means being able to chose two digits, and having no choice for the third one. But those two digits have to be of the same parity, otherwise their average would not be an integer. Indeed, take $i,j \in \mathbb{N}$ , then $(2i+2j)/2$ and $(2i+1+2j+1)/2$ are in $\mathbb{N}$ whereas $(2i+1+2j)/2$ is not.

We also have to make sure that the first digit is positive. Therefore we count : For $(b,c)$ both pair numbers we have $5\times 5$ choices minus 1 (we exclude $(0,0)$ ) and for $(a,c)$ or $(a,b)$ pairs we have $4 \times 5$ choices each because $a$ is chosen in $\{2,4,6,8\}$ and $b \in \{0,2,4,6,8\}$ . But $222,444,666,888$ are counted twice too many. Adding all this up gives : $5^2 - 1 + 2\times 4 \times 5 - 2 \times 4 = 56$

For impairs we don't have the problem with $a$ being equal to $0$ , so $a,b,c$ all play the same role. For each pair of two digits $(a,b)$ , $(b,c)$ , or $(a,c)$ we can chose $5\times 5$ combinations, and removing the $111,333,555,777,999$ twice gives $5^2 \times 3 - 5\times 2 = 65$ .

Finally, the result is $65 + 56 = 121$

With one of the digit as $1$ to begin with:

102, 111 $\rightarrow$ possible cases

102 can be permuted in 4 ways to give a three digit number. 111 can permuted in 1 way for the same reason.

Now, let a digit be $2$

240, 231, 222 $\rightarrow$ possible cases

240 $\rightarrow$ 4 sub-cases

231 $\rightarrow$ 6 sub-cases

222 $\rightarrow$ 1 sub-case

Now let a digit be $3$

360, 351, 342, 333 $\rightarrow$ possible cases

360 $\rightarrow$ 4 sub-cases

351 $\rightarrow$ 6 sub-cases

342 $\rightarrow$ 6 sub-cases

333 $\rightarrow$ 1 sub-case

Now let a digit be $4$

408, 417, 426, 453, 444 $\rightarrow$ possible cases

408 $\rightarrow$ 4 sub-cases

417 $\rightarrow$ 6 sub-cases

426 $\rightarrow$ 6 sub-cases

453 $\rightarrow$ 6 sub-cases

444 $\rightarrow$ 1 sub-case

Now let a digit be $5$

519, 528, 537, 546, 555 $\rightarrow$ possible cases

519 $\rightarrow$ 6 sub-cases

528 $\rightarrow$ 6 sub-cases

537 $\rightarrow$ 6 sub-cases

546 $\rightarrow$ 6 sub-cases

555 $\rightarrow$ 1 sub-case

Now let a digit be $6$

639, 648, 657, 666 $\rightarrow$ possible cases

639 $\rightarrow$ 6 sub-cases

648 $\rightarrow$ 6 sub-cases

657 $\rightarrow$ 6 sub-cases

666 $\rightarrow$ 1 sub-case

Now let a digit be $7$

759, 768, 777 $\rightarrow$ possible cases

759 $\rightarrow$ 6 sub-cases

768 $\rightarrow$ 6 sub-cases

777 $\rightarrow$ 1 sub-case

Now let a digit be $8$

879, 888 $\rightarrow$ possible cases

879 $\rightarrow$ 6 sub-cases

888 $\rightarrow$ 1 sub-case

At last, let a digit be $9$

999 is the only case and sub-case.

Now sum of all the subcases is $\boxed{121}$

We can create 3 cases as: 111,222=>9 numbers

AP triplets having 0 =>(0,1,2),(0,2,4),(0,3,6),(0,4,8) 4 each

AP triplets without 0 (6 numbers each)=>1(common difference=4)+3(c.d=3)+5(c.d=2)+7(c.d=1)=16

9+16+96=121