23. Three balls are randomly and independently thrown into buckets numbered with positive integers, such that the probability of each ball being thrown into bucket $i (i=1,2, \cdots)$ is $2^{-i}$, and no bucket can contain more than one ball. The probability that the three balls land in buckets with numbers forming an arithmetic sequence is $\frac{p}{q}$ (where $p$ and $q$ are coprime positive integers). Then $p+q=$ ( ).
(A) 55
(B) 56
(C) 57
(D) 58
(E) 59
23. A.

Assuming the middle bucket is labeled as $x$, then the first bucket has $x-1$ different possibilities.

By the property of arithmetic sequences, the label of the last bucket is uniquely determined by the labels of the first two buckets.

Given that the probability of selecting three buckets is $6 \times 2^{-3 x}$, the probability of selecting a bucket labeled $x$ is $\frac{6(x-1)}{8^{x}}$.
$$
\begin{array}{l} 
\text { Therefore, } 6 \sum_{x=2}^{\infty} \frac{x-1}{8^{x}}=\frac{6}{8}\left(\frac{1}{8}+\frac{2}{8^{2}}+\frac{3}{8^{3}}+\cdots\right) \\
= \frac{3}{4}\left(\left(\frac{1}{8}+\frac{2}{8^{2}}+\frac{3}{8^{3}}+\cdots\right)+\right. \\
\left.\left(\frac{1}{8^{2}}+\frac{2}{8^{3}}+\frac{3}{8^{4}}+\cdots\right)+\cdots\right) \\
= \frac{3}{4}\left(\frac{1}{7}\left(1+\frac{1}{8}+\frac{1}{8^{2}}+\frac{1}{8^{3}}+\cdots\right)\right) \\
= \frac{3}{4} \times \frac{8}{49}=\frac{6}{49} .
\end{array}
$$

Therefore, $p+q=6+49=55$.