Example 1 Let $\left\{a_{n}\right\}$ be an arithmetic sequence with the first term $a_{1}$ and common difference $d$. The sequence $\left\{a_{n}\right\}$ is grouped according to the rule of $3k$ numbers in the $k$-th group as follows:
$$
\left(a_{1}, a_{2}, a_{3}\right),\left(a_{4}, a_{5}, \cdots, a_{9}\right),\left(a_{10}, a_{11}, \cdots, a_{18}\right),
$$
$\cdots$.

Determine which group and which number in the group $a_{n}$ is. Also, find the sum of the terms in the group where $a_{n}$ is located.
Explanation: If $a_{n}$ is in the $k$-th group, then the first $k-1$ groups have
$$
3+6+9+\cdots+3(k-1)=\frac{3 k(k-1)}{2}
$$

terms. Therefore,
$$
\frac{3 k(k-1)}{2}+1 \leqslant n0 .\end{array}\right.$$
Solving this system of equations yields
$$
-\frac{1}{2}+\frac{\sqrt{24 n-15}}{6}<k \leqslant \frac{1}{2}+\frac{\sqrt{24 n-15}}{6} \text {. }
$$

Since $k \in \mathbf{N}_{+}$, and
$$
\frac{1}{2}+\frac{\sqrt{24 n-15}}{6}-\left(-\frac{1}{2}+\frac{\sqrt{24 n-15}}{6}\right)=1 \text {, }
$$

we have $k=\left[\frac{1}{2}+\frac{\sqrt{24 n-15}}{6}\right]$.
Thus, $a_{n}$ is the $n-\frac{3 k(k-1)}{2}$-th term in the $k$-th group, where $k=\left[\frac{1}{2}+\frac{\sqrt{24 n-15}}{6}\right]$.
Since the $k$-th group is an arithmetic sequence with
$$
a_{\frac{3 k(k-1)}{2}+1}=a_{1}+\frac{3 k(k-1)}{2} d
$$

as the first term, $d$ as the common difference, and a total of $3 k$ terms, the sum of all its terms is
$$
\begin{array}{l}
k\left(a_{1}+\frac{3 k(k-1)}{2} d\right)+\frac{3 k(3 k-1)}{2} d \\
=k a_{1}+\frac{3 k\left(k^{2}+2 k-1\right) d}{2},
\end{array}
$$

where $k=\left[\frac{1}{2}+\frac{\sqrt{24 n-15}}{6}\right]$.
Note: Grouped sequence problems are sometimes given in the form of a number table. In such cases, converting the problem into a grouped sequence problem can solve it.