Answer
$a_n=\dfrac{7}{3}-\dfrac{1}{3}n$ ; $a_{20}=\dfrac{-13}{3}$
Work Step by Step
We know that $a_n= a_1+(n-1) d$
Here, we have $d=\dfrac{5}{3}-2=-\dfrac{1}{3}$
$a_n= 2+(n-1) \times (\dfrac{-1}{3})$
This gives: $a_n=\dfrac{7}{3}-\dfrac{1}{3}n$
$a_{20}=\dfrac{7}{3}-\dfrac{1}{3} \times 20=\dfrac{-13}{3}$
Hence, $a_n=\dfrac{7}{3}-\dfrac{1}{3}n$ ; $a_{20}=\dfrac{-13}{3}$
