Example 1. For the arithmetic sequence $\{3 n-1\}$, if three arithmetic means are inserted between each pair of consecutive terms, forming a new sequence. What is the position of the 12th term of the original sequence in the new sequence? What is the position of the 29th term of the new sequence in the original sequence?
Solve: The general term of the original sequence is $a_{n}=3(n-1)+2$, then the general term of the new sequence is $b_{m}=\frac{3}{4}(m-1)+2$.

We find that $m=4 n-3$, so $a_{12}=b_{45}$, $b_{21}=a_{8}$.