3. In the sequence $\left\{a_{n}\right\}$, adjacent terms $a_{n} 、 a_{n+1}$ are the roots of the equation $x^{2}+3 n x+b_{n}=0$. It is known that $a_{11}=-17$. Then the value of $b_{s 1}$ is ( ).
(A) 5800
(B) 5840
(C) 5860
(D) 6000
3. B.

Given that $a_{n}+a_{n+1}=-3 n$. Then
$$
\begin{array}{l}
a_{n+2}-a_{n}=\left(a_{n+2}+a_{n+1}\right)-\left(a_{n+1}+a_{n}\right) \\
=-3(n+1)-(-3 n)=-3 .
\end{array}
$$

Therefore, $a_{1}, a_{3}, \cdots, a_{2 n+1}$ and $a_{2}, a_{4}, \cdots, a_{2 n}$ are both arithmetic sequences with a common difference of -3. Hence
$$
\begin{array}{l}
a_{52}=a_{10}+21 \times(-3)=-80, \\
a_{51}=a_{11}+20 \times(-3) .
\end{array}
$$

Also, $a_{10}+a_{11}=-30$, so $a_{11}=-13$.
Thus, $a_{51}=-73, b_{51}=a_{51} a_{52}=5840$.