smc

Note that the given sequence is a geometric sequence with a common ratio $10^{\frac{1}{11}}$. Let the product of the first $n$ terms of the sequence be denoted $P_n$. It is a consequence of the laws of exponents that $P_1=(10^{\frac{1}{11}}{)}^1$, $P_2=(10^{\frac{1}{11}}{)}^{1+2}$, and, in general, $P_n=(10^{\frac{1}{11}}{)}^{T_n}$, where $T_n$ denotes the $n$th triangular number. Setting $P_n$ equal to $100,000$, we see that: $$\begin{array}{rl}& \\ (10^{\frac{1}{11}}{)}^{\frac{n(n+1)}{2}}& =10^5\\ \frac{1}{11}\ast \frac{n(n+1)}{2}& =5\\ n(n+1)& =110\\ n^2+n-110& =0\\ (n-10)(n+11)& =0\end{array}$$ Because $n$ must be positive, we are left with $n=10$. Given this information, choice (D) may seem appealing. Do not be fooled. The product of the first $n$ terms is exactly $100,000$, but the problem asks for the smallest $n$ such that $P_n$ exceeds 100,000. Thus, the minimum value of $n$ which satsifies the problem is $\boxed{11}$.