Answer
$a_n=-\dfrac{2}{3}+\dfrac{2}{3}n$ ; $a_{20}=\dfrac{38}{3}$
Work Step by Step
We know that $a_n= a_1+(n-1) d$
Here, we have $d=\dfrac{2}{3}-0=\dfrac{2}{3}$
$a_n= 0+(n-1) \times (\dfrac{2}{3})$
This gives: $a_n=-\dfrac{2}{3}+\dfrac{2}{3}n$
$a_{20}=-\dfrac{2}{3}+\dfrac{2}{3} \times 20=\dfrac{38}{3}$
Hence, $a_n=-\dfrac{2}{3}+\dfrac{2}{3}n$ ; $a_{20}=\dfrac{38}{3}$
