Question

Find sum of first $18$ term for an A.P $\frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{10}{3},.....$

Answer 1

Given $AP=\frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{10}{3},......$

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

$d=\frac{4}{3}-\frac{1}{3}=1$

$n=18$

Thus,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{18}{2}[2\times \frac{1}{3}+(18-1\left)1\right]$

$=9[\frac{2}{3}+17]$

$=9\left[\frac{2+51}{3}\right]$

$=3\times 53$

$=159$