Given the sequence:

$$
\(\frac{7}{3}, \frac{7}{4}, \frac{7}{5}, \ldots\)
$$


We want to find the number $n$ such that:

$$
\(a_n=a_6-a_7\)
$$

where $a_n$ is the $\(n^{\text {th }}\)$ term of the sequence.
Step 1: Identify the $\(n^{\text {th }}\)$ term
The general term of the sequence is:

$$
\(a_n=\frac{7}{n+2}\)
$$

because when $n=1$, term is $\(\frac{7}{3}\)$, so the denominator is $\(n+2\)$.
Step 2: Calculate $\(a_6\)$ and $\(a_7\)$

$$
\(\begin{aligned}
& a_6=\frac{7}{6+2}=\frac{7}{8}, \\
& a_7=\frac{7}{7+2}=\frac{7}{9} .
\end{aligned}\)
$$


Step 3: Find $\(a_6-a_7\)$

$$
\(a_6-a_7=\frac{7}{8}-\frac{7}{9}=7\left(\frac{1}{8}-\frac{1}{9}\right)=7\left(\frac{9-8}{72}\right)=\frac{7}{72}\) .
$$

Step 4: Find $n$ such that $\(a_n=\frac{7}{72}\)$
Recall $\(a_n=\frac{7}{n+2}\)$, so:

$$
\(\frac{7}{n+2}=\frac{7}{72} \Longrightarrow n+2=72 \Longrightarrow n=70 \text {. }\)
$$


Final answer:

$$
\(a_n=a_6-a_7 \text { when } n=70\)
$$


So, the answer is:
D. $\(70^{\text {th }}\)$ term.