55 In a geometric sequence $\left\{a_{n}\right\}$, the first term $a_{1}=2^{-5}$, and the geometric mean of its first 11 terms is $2^{5}$. If, among the first 11 terms, one term is removed and the geometric mean of the remaining terms is $2^{4}$, then the removed term is
A. The 8th term
B. The 9th term
C. The 10th term
D. The 11th term
55 D. Let the common ratio of the geometric sequence be $q$, and the term removed is the $k$-th term, then $a_{k}=\left(2^{5}\right)^{11} \div\left(2^{4}\right)^{10}=2^{15}$. Also, $2^{-5} \cdot q^{(1+2+\cdots+10) / 11}=2^{5}$, so $q=2^{2}$. From $2^{15}=\left(2^{-5}\right) \cdot\left(2^{2}\right)^{10}=a_{1} \cdot q^{10}$ we know $k=11$.