Answer
$a_n=0.5(n-1)$
and $a_{51} = 25$
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: $a_1=0 \\ d=\dfrac{1}{2}=0.5$
Next, we will substitute the above data into formula (1) to obtain:
$a_n=0+0.5(n-1) \implies a_n=0.5(n-1)$
In order to compute the $51st$ term, we need to plug in $51$ for $n$ into the above form to obtain:
$a_{51} = 0.5(51-1) =0.5(50)$
So, $a_{51} = 25$
