1. Given an arithmetic sequence $\left\{a_{n}\right\}$ with common difference $d \neq 0$, and $a_{5}^{2}+a_{7}^{2}+16 d=a_{9}^{2}+a_{11}^{2}$, then the sum of the first 15 terms of $\left\{a_{n}\right\}$, $S_{15}$, is
A. 15
B. 16
C. 30
D. 32
$$
\begin{array}{l}
\left(a_{1}+4 d\right)^{2}+\left(a_{1}+6 d\right)^{2}+16 d=\left(a_{1}+8 d\right)^{2}+\left(a_{1}+10 d\right)^{2} \\
\Rightarrow 20 a_{1} d+52 d^{2}+16 d=36 a_{1} d+164 d^{2} \Rightarrow 16 a_{1}+112 d=16 \Rightarrow a_{8}=1 .
\end{array}
$$

Therefore, \( S_{15}=15 a_{8}=15 \), so the answer is \( A \).