Find the nth term of the sequence, then find the 20th term, 100th term and the series for each? 2/5, 1/15, -4/15,....

[tex]a_1=\dfrac{2}{5}=\dfrac{6}{15}\\\\a_2=\dfrac{1}{15}\\\\a_3=-\dfrac{4}{15}\\\\a_2-a_1=\dfrac{1}{15}-\dfrac{6}{15}=-\dfrac{5}{15}=-\dfrac{1}{3}\\\\a_3-a_2=-\dfrac{4}{15}-\dfrac{1}{15}=-\dfrac{5}{15}=-\dfrac{1}{3}[/tex]

It's an arithmetic sequence where

[tex]a_1=\dfrac{2}{5}\ and\ d=-\dfrac{1}{3}[/tex]


nth term of the arithmetic sequence is equal:

[tex]a_n=a_1+(n-1)d[/tex]

Substitute:

[tex]a_n=\dfrac{2}{5}+(n-1)\cdot\left(-\dfrac{1}{3}\right)\\\\a_n=\dfrac{2}{5}-\dfrac{1}{3}n+\dfrac{1}{3}\\\\a_n=\dfrac{6}{15}+\dfrac{5}{15}-\dfrac{1}{3}n\\\\\boxed{a_n=\dfrac{11}{15}-\dfrac{1}{3}n}[/tex]

Other form:

[tex]a_n=\dfrac{11}{15}-\dfrac{5}{15}n\\\\\boxed{a_n=\dfrac{11-5n}{15}}[/tex]


[tex]a_{20}=\dfrac{11-5\cdot20}{15}=\dfrac{11-100}{15}=\dfrac{-89}{15}\\\\a_{100}=\dfrac{11-5\cdot100}{15}=\dfrac{11-500}{15}=\dfrac{-489}{15}=-\dfrac{163}{5}[/tex]