Correct Answer: $\frac{3816}{325}$

Solution : Given: The series is $\frac{7}{2}+\frac{11}{3}+\frac{7}{6}+\frac{11}{15}+\frac{7}{12}+\frac{11}{35}+\ldots \ldots+\frac{7}{156}+\frac{11}{575}$.
Determine the value of the series by solving two distinct series independently.
The sum of the first series = $\frac{7}{2}+\frac{7}{6}+\frac{7}{12}+...+\frac{7}{156}$
⇒ Sum = $7[1–(\frac{1}{2})+(\frac{1}{2}–\frac{1}{3})+(\frac{1}{3}–\frac{1}{4})+...+(\frac{1}{12}–\frac{1}{13})]$
⇒ Sum = $7[1–\frac{1}{13}]$
⇒ Sum = $7\times\frac{12}{13}=\frac{84}{13}$
The sum of the second series = $\frac{11}{3}+\frac{11}{15}+\frac{11}{35}+...+\frac{11}{575}$
⇒ Sum = $11[\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{575}]$
⇒ Sum = $\frac{11}{2}[\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{575}]$
⇒ Sum = $\frac{11}{2}[1–(\frac{1}{3})+(\frac{1}{3}–\frac{1}{5})+(\frac{1}{5}–\frac{1}{7})+...+(\frac{1}{23}–\frac{1}{25})]$
⇒ Sum = $\frac{11}{2}[1–\frac{1}{25}]$
⇒ Sum = $\frac{11}{2}\times\frac{24}{25}=\frac{132}{25}$
The total sum of the series = The sum of the first series + The sum of the second series
The total sum of the series $=\frac{84}{13}+\frac{132}{25}$
$=\frac{84\times 25+132\times 13}{25\times 13}=\frac{2100+1716}{325}=\frac{3816}{325}$
Hence, the correct answer is $\frac{3816}{325}$.