Example 1 Given that the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both arithmetic sequences, $S_{n}$ and $T_{n}$ are their respective sums of the first $n$ terms, and $\frac{S_{n}}{T_{n}}=\frac{7 n+1}{n+3}$, find the value of $\frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}$.
Analysis: Students generally can use the property of the general term of an arithmetic sequence: "If $m, n, k, l \in \mathbf{N}^{*}$ and $m+n=k+l$, then $a_{m}+a_{n}=a_{k}+a_{l}$" to simplify the expression $\frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}$. However, students often fail to establish the relationship between the sum of the first $n$ terms and the general term formula of an arithmetic sequence, leading to cumbersome calculations. In fact, it is only necessary to constructively use $S_{n}=\frac{\left(a_{1}+a_{n}\right) n}{2}$.
$$\begin{array}{l}
\because \frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}=\frac{\left(a_{5}+a_{17}\right)+\left(a_{2}+a_{22}\right)}{\left(b_{10}+b_{12}\right)+\left(b_{8}+b_{16}\right)} \\
=\frac{2 a_{11}+2 a_{12}}{2 b_{11}+2 b_{12}}=\frac{\frac{22\left(a_{12}+a_{12}\right)}{2}}{\frac{22\left(b_{11}+b_{12}\right)}{2}}=\frac{\frac{22\left(a_{1}+a_{22}\right)}{2}}{\frac{22\left(b_{1}+b_{22}\right)}{2}}=\frac{S_{22}}{T_{22}}, \\
\therefore \frac{a_{2}+a_{5}+a_{17}+a_{22}}{b_{8}+b_{10}+b_{12}+b_{16}}=\frac{S_{22}}{T_{22}}=\frac{7 \times 22+1}{22+3}=\frac{31}{5} \text{. }
\end{array}$$