Let the digits a, b, c be in A.P. Nine-digit numbers are to be...

Three numbers $a, b, c$ are in A.P. ad they are used to making $9 -$ digit number using each digit thrice such that atleast three consecutive digits are in A.P., So, $c, b, a$ will also be in A.P.

Now, we have $7$ position to put \left(a, b, c\right)or \left(c, b, a\right)as \mathop{\mathop{_______}_{⏟}}_{7positions}__.

Number of such numbers is

$= C1 2 \cdot C1 7 \cdot$\dfrac{6 !}{2 ! \cdot 2 ! \cdot 2 !}

$= 1260$