Let the digits a, b, c be in A. P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?

<p>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.</p> <p>We have the two possible sequences for the AP :</p> <ol> <li>a, b, c</li> <li>c, b, a</li> </ol> <p>This shows the flexibility in ordering the three digits that are in AP in our nine-digit number.</p> <p>The next step is to choose the location of this sequence of three numbers within our nine-digit number.</p> <p>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.</p> <p>Therefore, the number of ways to select 3 consecutive places out of the 9 places for the AP sequence is 7.</p> <p>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.</p> <p>The remaining 6 digits (two 'a', two 'b', two 'c") can be arranged in $\frac{6!}{2!2!2!}$ ways.</p> <p>Therefore, the total number of such nine-digit numbers is $^7C_1 \times 2 \times \frac{6!}{2!2!2!} = 1260$.</p>