16.2.1 ${ }^{\star}$ A four-digit number, the digits in the thousands, hundreds, tens, and units place form an arithmetic sequence. The digit in the hundreds place is the geometric mean of the digits in the units and thousands places. The sum of the number formed by arranging these four digits in reverse order and the original four-digit number is 11110. Find the original four-digit number.
Let the original four-digit number be $\overline{a b c d}$. Then, from $\overline{a b c d} + \overline{d c b a} = 11110$, we get
$$
b+c=a+d=10.
$$

The common difference of the sequence $a, b, c, d$ can only be $0, 1, 2$. If the common difference is 0, then by (1), $a=b=c=d=5$. If the common difference is 2, then by (1) we get $b+1=5$ or $c+1=5$, thus $a=2, b=4, c=6, d=8$ or $a=8, b=6, c=4, d=2$, but the latter does not satisfy the condition $b^2 = ad$. A common difference of 1 cannot occur because $b+c$ is even.
Therefore, the original four-digit number is 5555 or 2468.