Question

Find the sum of the following A.P
$\frac{1}{15},\frac{1}{12},\frac{1}{10},\dots \dots ,\text{to}11\text{terms}$

Answer 1

Given A.P is $\frac{1}{15},\frac{1}{12},\frac{1}{10},\dots \dots$ to $11$ terms
For this A.P.,
First term is $a=\frac{1}{15}$
Number of terms is $n=11$
Common difference is $d=a_2-a_1=\frac{1}{12}-\frac{1}{15}$
$\Rightarrow d=\frac{5-4}{60}=\frac{1}{60}$

Since, we have $S_n=\frac{n}{2}[2\left(a\right)+(n-1\left)d\right]$
$S_{11}=\frac{11}{2}[2\left(\frac{1}{15}\right)+(11-1\left)\frac{1}{60}\right]$
$=\frac{11}{2}[\frac{2}{15}+\frac{10}{60}]$
$=\frac{11}{2}[\frac{2}{15}+\frac{1}{6}]$
$=\frac{11}{2}\left[\frac{4+5}{30}\right]$
$=\left(\frac{11}{2}\right)\left(\frac{9}{30}\right)\therefore S_{11}=\frac{33}{20}$

Hence, the sum of the terms given A.P is $\frac{33}{20}$