Answer
$a_n=1-\frac{1}{3}(n-1)$
$a_{51}=-\frac{47}{3}$
Work Step by Step
RECALL:
The $n^{th}$ term of an arithmetic sequence can be found using the formula:
$a_n = a_1 + (n-1)d$
where $a_1$ = first term and $d$ = common difference
The given sequence has:
$a_1=1$;
$d=-\frac{1}{3}$
Substitute these values into the formula for the $n^{th}$ term to obtain:
$a_n=a_1 + (n-1)d
\\a_n=1+(n-1)(-\frac{1}{3})
\\a_n=1+(-\frac{1}{3})(n-1)
\\a_n=1-\frac{1}{3}(n-1)$
To find the 51st term, substitute $51$ for $n$ to obtain:
$a_{51} = 1-\frac{1}{3}(51-1)
\\a_{51}=1-\frac{1}{3}(50)
\\a_{51} = 1-\frac{50}{3}
\\a_{51} = \frac{3}{3}-\frac{50}{3}
\\a_{51}=-\frac{47}{3}$