For the sequence $\left\{a_{n}\right\}$, let $b_{n}=a_{n+1}-a_{n}$, then the sequence $\left\{b_{n}\right\}$ is the first-order difference sequence of $\left\{a_{n}\right\}$, and let $c_{n}=b_{n+1}-b_{n}$, then the sequence $\left\{c_{n}\right\}$ is the second-order difference sequence of $\left\{a_{n}\right\}$. By analogy, the $p$-th order difference sequence of $\left\{a_{n}\right\}$ can be obtained. If the $p$-th order difference sequence of $\left\{a_{n}\right\}$ is a non-zero constant sequence, then the sequence $\left\{a_{n}\right\}$ is called a $p$-th order arithmetic sequence. Given that the sequence $\left\{a_{n}\right\}$ is a second-order arithmetic sequence, $a_{10}=23, a_{20}=23$, and in the second-order difference sequence $\left\{c_{n}\right\}$, $c_{n}=20$, then $a_{30}=$ $\qquad$ .
$$
\begin{array}{l}
a_{n}=x n^{2}+y n+z, b_{n}=a_{n+1}-a_{n}=x\left[(n+1)^{2}-n^{2}\right]+y[(n+1)-n] \\
=x(2 n+1)+y=2 x n+x+y, c_{n}=b_{n+1}-b_{n}=2 x=20 \Rightarrow x=10 . \\
\text { Solving the system }\left\{\begin{array} { l } 
{ a _ { 1 0 } = 1 0 \cdot 1 0 0 + 1 0 y + z = 2 3 , } \\
{ a _ { 2 0 } = 1 0 \cdot 4 0 0 + 2 0 y + z = 2 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
y=-300, \\
z=2023 .
\end{array}\right.\right.
\end{array}
$$

Therefore, $a_{n}=10 n^{2}-300 n+2023 \Rightarrow a_{30}=2023$.