Question

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

Answer 1

The first term is $\frac{1}{15}$ and the common difference is $\frac{1}{12}-\frac{1}{15}=\frac{1}{60}$.

The sum of the A.P up to 11 terms is,

$S_{11}=\frac{11}{2}\left[2\left(\frac{1}{15}\right)+\left(11-1\right)\times \frac{1}{60}\right]$

$=\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right]$

$=\frac{11}{2}\left(\frac{3}{10}\right)$

$=\frac{33}{20}$