Answer
$a_n=6-2(n-1)$
and $a_{51} =-94$
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:
$a_1=6 \\ d= -2$
Next, we will substitute the above data into formula (1) to obtain:
$a_n=6+(-2)(n-1) \implies a_n=6-2(n-1)$
In order to compute the 51st term, we need to plug in $51$ for $n$ into the above form to obtain:
$a_{51} = 6-2(51-1) =6-100$
So, $a_{51} =-94$
