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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Updated on: 10-Oct-2022

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