Find the sum of the following APs:
$\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, ……..,$ to $11$ terms.
 Given:
Given AP is $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, ……..,$
To do:
We have to find the sum of the given A.P. to 11 terms.
Solution:
$a=\frac{1}{15}, d=\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}, n=11$
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$S_{11}=\frac{11}{2}[2 \times \frac{1}{15}+(11-1) \frac{1}{60}]$
$=\frac{11}{2}[\frac{2}{15}+\frac{10}{60}]$
$=\frac{11}{2}(\frac{8+10}{60})$
$=\frac{11}{2} \times \frac{18}{60}$
$=\frac{33}{20}$
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