Correct Answer: $\frac{95120}{33}$

Solution : Given: The series is $99 \frac{11}{99}+99 \frac{13}{99}+99 \frac{15}{99}+....+99 \frac{67}{99}$.
⇒ $(99+99+99+......)+(\frac{11}{99}+\frac{13}{99}+...+\frac{67}{99})$
The number of terms in the series is given by $t_n=a+(n–1)d$ where $t_n$ is the last term, $a$ is the first term, $d$ is the common difference and $n$ is the number of terms.
⇒ $\frac{67}{99}=\frac{11}{99}+(n–1)\times\frac{2}{99}$
⇒ $67=11+(n–1)\times 2$
⇒ $56=2\times (n–1)$
⇒ $ (n–1)=28$
⇒ $n=29$
So, the number of terms is 29.
The given series can be written as $99\times 29+(\frac{11}{99}+\frac{13}{99}+...+\frac{67}{99})$.
$S_n=\frac{n}{2}[2a+(n–1)d]$
= $99\times 29+\frac{29}{2}\times[2\times\frac{11}{99}+28\times\frac{2}{99}]$
= $99\times 29+\frac{29\times11}{99}+29\times\frac{28}{99}=99\times 29+29(\frac{11}{99}+\frac{28}{99})$
= $99\times 29+29\times\frac{11+28}{99}=99\times 29+29\times\frac{39}{99}$
= $29[99+\frac{13}{33}]=29\times\frac{3267+13}{33}$
= $29\times\frac{3280}{33}=\frac{95120}{33}$
Hence, the correct answer is $\frac{95120}{33}$.