Question

Find the sum of first $18$ terms of the A.P. $\frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{10}{3},...$

Answer 1

$A.P:\frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{10}{3},.....$

$a=\frac{1}{3}$,

$d=\frac{4}{3}-\frac{1}{3}$ $=\frac{3}{3}$ $=1,n=18$

$S_n$=$\frac{n}{2}(2a+(n-1\left)d\right)$ $\Rightarrow S_{18}=\frac{18}{2}\left(2\right(\frac{1}{3})+(18-1\left)1\right)$

$S_{18}$=$9(\frac{2}{3}+17)$

$S_{18}$= 9 $\left(\frac{2+51}{3}\right)$

$S_{18}$=3$\times$53

$S_{18}$=159