Let the three digits be $a,b,c$ and they are in A.P.
So we may write them as three distinct digits with $b-a=c-b=d$ for some $d\neq 0$.

Thus any three consecutive digits are in A.P. iff they are one of the permutations of $(a,b,c)$? No — since for three terms $x,y,z$ to be in A.P., we need $2y=x+z.$ Because $a,b,c$ themselves are in A.P. with middle term $b$, the only 3-digit blocks formed from $a,b,c$ that are in A.P. are $(a,b,c)\quad \text{and} \quad (c,b,a).$ Indeed, the middle digit must be $b$, and the ends must be $a,c$ in either order.

So the problem becomes:

Count 9-digit arrangements of the multiset $\{a,a,a,b,b,b,c,c,c\}$ such that at least one occurrence of the consecutive block $abc$ or $cba$ appears.

1. Total number of arrangements

The total number of 9-digit arrangements using each of $a,b,c$ three times is $\frac{9!}{3!3!3!}=1680.$

We now count those with at least one A.P. block.

2. Count arrangements containing the block $abc$

Treat one occurrence of $abc$ as a single object. Then along with the remaining digits $a,a,b,b,c,c$, we have total objects:

• one block $[abc]$
• two more $a$'s
• two more $b$'s
• two more $c$'s

So the number of arrangements is $\frac{7!}{2!2!2!}=630.$

Similarly, the number containing the block $cba$ is also $630.$

If we add these, we get $1260$, but this double-counts arrangements containing both $abc$ and $cba$.

So we must subtract the intersection.

3. Count arrangements containing both $abc$ and $cba$

We need arrangements of the multiset with at least one $abc$ and at least one $cba$ as consecutive 3-blocks.

Treat these two blocks as objects: $[abc],\ [cba]$ The remaining digits are one $a$, one $b$, one $c$.

So naively we arrange 5 objects: $[abc],[cba],a,b,c$ which gives $\frac{5!}{1!1!1!}=120$ if all objects were distinct. But this naive count is actually valid because the leftover $a,b,c$ are distinct from the blocks.

However, we must check whether overlapping occurrences can happen and whether they are included correctly.

Overlap check

Can $abc$ and $cba$ overlap?
A 2-letter overlap would require suffix of one to equal prefix of the other:

• suffix of $abc$ of length 2 is $bc$, prefix of $cba$ of length 2 is $cb$ — not equal.
• suffix of $cba$ of length 2 is $ba$, prefix of $abc$ of length 2 is $ab$ — not equal.

A 1-letter overlap also does not create a valid simultaneous placement because end/start letters do not match suitably for distinct blocks.

So the two blocks must be disjoint. Hence the count $120$ is correct.

Thus $|abc\cap cba|=120.$

4. Apply inclusion-exclusion

Therefore the required number is $630+630-120=1140.$

5. Final answer

$\boxed{1140}$

6. Comparison with stored answer

Stored correct answer = $1260$.

But $1260$ equals $630+630$ and ignores the overlap cases containing both $abc$ and $cba$. Since $120$ such arrangements exist, the correct count should be $1260-120=1140.$

Hence I do not agree with the stored answer.