Find the sum of the following $APs$: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots,$ to $11$ terms.

(D) For the given $A.P.$,the first term $a = \frac{1}{15}$ and the number of terms $n = 11$.
The common difference $d$ is calculated as:
$d = a_2 - a_1 = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$.
The sum of $n$ terms of an $A.P.$ is given by the formula:
$S_n = \frac{n}{2} [2a + (n - 1)d]$.
Substituting the values $n = 11$,$a = \frac{1}{15}$,and $d = \frac{1}{60}$:
$S_{11} = \frac{11}{2} [2(\frac{1}{15}) + (11 - 1)(\frac{1}{60})]$.
$S_{11} = \frac{11}{2} [\frac{2}{15} + 10(\frac{1}{60})]$.
$S_{11} = \frac{11}{2} [\frac{2}{15} + \frac{1}{6}]$.
Taking the $LCM$ of $15$ and $6$,which is $30$:
$S_{11} = \frac{11}{2} [\frac{4 + 5}{30}] = \frac{11}{2} [\frac{9}{30}] = \frac{11}{2} [\frac{3}{10}] = \frac{33}{20}$.