Question

For A.P.$\frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{10}{3}.......$Find $T_{18}$.

Answer 1

$T_n=a+(n-1)d$

$T_{18}=a+(18-1)d$

$T_{18}=a+17d$ (1)

For the given A.P. $a=\frac{1}{3}$ (2)

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

Substituting (2) and (3) in (1)

$T_{18}=\frac{1}{3}+17\left(1\right)$

$T_{18}=\frac{1}{3}+17$

$T_{18}=\frac{52}{3}$