Answer
$a_n=\dfrac{4-n}{3}$
$a_{51}=-\dfrac{47}{3}$
Work Step by Step
We know that the $n^{th}$ term of an aritihmetic sequence is given by the formula
$a_n=a_1+(n-1)d$
where $d$=common difference and $a_1$= the first term
Hence, here we have
$a_n=1+(n-1)\cdot\left(-\dfrac{1}{3}\right)\\a_n=1+\left(-\dfrac{n}{3}\right)+\dfrac{1}{3}\\a_n=\dfrac{4}{3}-\dfrac{n}{3}\\a_n=\dfrac{4-n}{3}$
Therefore,
$a_{51}=\dfrac{4-51}{3}\\a_{51}=-\dfrac{47}{3}$
