In the sequence of fractions $\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}, \frac{5}{1}, \ldots$, fractions equivalent to any given fraction occur many times. For example, fractions equivalent to $\frac{1}{2}$ occur for the first two times in positions 3 and 14. In which position is the fifth occurrence of a fraction equivalent to $\frac{3}{7}$ ?
(A) 1207
(B) 1208
(C) 1209
(D) 1210
(E) 1211
In analyzing this sequence of fractions, we start by observing that this large sequence is itself made up of smaller sequences. Each of these smaller sequences is of the form

$$
\frac{n}{1}, \frac{n-1}{2}, \frac{n-2}{3}, \ldots, \frac{3}{n-2}, \frac{2}{n-1}, \frac{1}{n}
$$

with the denominators increasing from 1 to $n$ and the numerators decreasing from $n$ to 1 .

We observe that there is 1 term in the first of these smaller sequences, 2 terms in the second of these, and so on. This can be seen in the following grouping:

$$
\left(\frac{1}{1}\right),\left(\frac{2}{1}, \frac{1}{2}\right),\left(\frac{3}{1}, \frac{2}{2} \frac{1}{3}\right),\left(\frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}\right), \ldots,\left(\frac{9}{1}, \frac{8}{2}, \frac{7}{3}, \frac{6}{4}, \frac{5}{5}, \frac{4}{6}, \frac{3}{7}, \frac{2}{8}, \frac{1}{9}\right), \ldots
$$

If we take any fraction in any of these smaller sequences, the sum of the numerator and denominator is 1 greater than the number of terms in this smaller sequence. For example, if we take the first occurrence of $\frac{3}{7}$, it would occur in the sequence with 9 terms.

This implies that the fifth occurrence of a fraction equivalent to $\frac{3}{7}$, namely $\frac{5 \times 3}{5 \times 7}=\frac{15}{35}$, would occur in the sequence with 49 terms, and would be the $35^{\text {th }}$ term in that sequence.

Since the smaller sequences before this particular sequence have $1,2,3, \ldots, 48$ terms, so the term $\frac{15}{35}$ is term number $(1+2+\cdots+47+48)+35=\frac{1}{2}(48)(49)+35=1176+35=1211$.

Answer: (E)