1. Given a sequence of numbers
$$
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \cdots, \frac{k}{1}, \frac{k-1}{2}, \cdots, \frac{1}{k} \text {. }
$$

In this sequence, the index of the 40th term that equals 1 is ( ).
(A) 3120
(B) 3121
(C) 3200
(D) 3201
- 1. B.

For the terms where the sum of the numerator and denominator is $k+1$, we denote them as the $k$-th group. According to the arrangement rule, the 40th term with a value of 1 should be the 40th number in the $2 \times 40-1=79$ group, with the sequence number being
$$
\begin{array}{l}
(1+2+\cdots+78)+40 \\
=\frac{(1+78) \times 78}{2}+40=3121 .
\end{array}
$$