10 . The sequence $\left\{3^{n-1}\right\}$ is grouped according to the rule “the $n$-th group has $n$ numbers” as follows:
$$
\text { (1),(3,9), }(27,81,243), \cdots \text {. }
$$

Then the first number of the 100th group is ( ).
(A) $3^{4950}$
(B) $3^{5000}$
(C) $3^{5010}$
(D) $3^{5050}$
10.A.

The sum of the first 99 terms is $1+2+\cdots+99=4950$. The 1st group is $3^{\circ}$, so the first number in the 100th group should be $3^{4950}$.