5. Given arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, let $A_{n}=a_{1}+a_{2}+\cdots+a_{n}, B_{n}=b_{1}+b_{2}+\cdots+b_{n}$. If for $\forall n \in \mathbf{N}^{*}$, $\frac{A_{n}}{B_{n}}=\frac{3 n+5}{5 n+3}$, then $\frac{a_{10}}{b_{6}}=(\quad)$
$\mathrm{A}: \frac{35}{33}$
$\mathrm{B}: \frac{31}{29}$
C: $\frac{175}{99}$
D: $\frac{155}{87}$
Answer $\mathrm{B}$.
Analysis Since $\left\{a_{n}\right\},\left\{b_{n}\right\}$ are arithmetic sequences, and the ratio of the sum of the first $n$ terms $\frac{A_{n}}{B_{n}}=\frac{3 n+5}{5 n+3}$, we can set
$$
A_{n}=k n(3 n+5), \quad B_{n}=k n(5 n+3)
$$

Thus
$$
\left\{\begin{array}{l}
a_{10}=A_{10}-A_{9}=10 k(30+5)-9 k(27+5)=62 k \\
b_{6}=B_{6}-B_{5}=6 k(30+3)-5 k(25+3)=58 k
\end{array} \quad \Rightarrow \frac{a_{10}}{b_{6}}=\frac{62 k}{58 k}=\frac{31}{29}\right.
$$