The problem involves forming nine-digit numbers from three digits a, b, c which are in Arithmetic Progression (AP), used three times each, such that at least once, three consecutive digits are in AP.

We have the two possible sequences for the AP :

1. a, b, c
2. c, b, a

This shows the flexibility in ordering the three digits that are in AP in our nine-digit number.

The next step is to choose the location of this sequence of three numbers within our nine-digit number.

Since there are nine places in our number and our sequence takes up three places, we have seven different starting points for our sequence : it can start at the first place, the second place, and so on, up to the seventh place.

Therefore, the number of ways to select 3 consecutive places out of the 9 places for the AP sequence is 7.

However, we also have to account for the fact that our sequence can be in one of two orders (a, b, c or c, b, a). So, we multiply the number of starting points by 2 to get $^7C_1 \times 2 = 14$ ways to arrange the sequence within our nine-digit number.

The remaining 6 digits (two 'a', two 'b', two 'c") can be arranged in $\frac{6!}{2!2!2!}$ ways.

Therefore, the total number of such nine-digit numbers is $^7C_1 \times 2 \times \frac{6!}{2!2!2!} = 1260$.