The sum of the 9th term and 12th term of the series \(\frac{1}{7}, \frac{1}{11}, \frac{1}{15}, \ldots\)…

The sum is $\frac{10}{221}$. The denominators form an arithmetic progression with first term 7 and common difference 4, so the ninth and twelfth terms are $\frac{1}{39}$ and $\frac{1}{51}$. Their sum is $\frac{51+39}{39\times51}$ $=\frac{90}{1989}$ $=\frac{10}{221}$.