Answer
$a_n=\dfrac{4-n}{3}$
and $a_{51} = \dfrac{-47}{3}$
Work Step by Step
The $n^{th}$ term of an arithmetic sequence is given by the formula:
$a_n = a_1 + (n-1)d (1)$
where
$a_1 = \ First \ Term; \\ d = \ Common \ Difference$
We have: $a_1=1 \\ d=\dfrac{-1}{3}$
Next, we will substitute the above data into formula (1) to obtain:
$a_n=1+(\dfrac{-1}{3})(n-1) \\ a_n=\dfrac{4}{3}-\dfrac{n}{3} \implies a_n=\dfrac{4-n}{3}$
In order to compute the $51st$ term, we need to plug in $51$ for $n$ into the above form to obtain:
$a_{51} = \dfrac{4-51}{3}$
So, $a_{51} = \dfrac{-47}{3}$
