Answer
First term: $a=195$
$n^{th}$ term: $a_n=195-7(n-1)$
Work Step by Step
RECALL:
The $n^{th}$ $a_n$ of an arithmetic sequence is given by the formula:
$a_n = a + d(n-1)$
where
$a$ = first term
$d$ = common difference
The twelfth term is $118$ . This means that:
$a_{12} = a + d(12-1)
\\118 = a + d(11)
\\118 = a +11d$
The eight term is $146$. This means that:
$a_8=a+d(8-1)
\\146=a + d(7)
\\146 = a + 7d$
The given information about the sequence results to two equations:
(equation 1) $118=a+11d$
(equation 2) $146=a+7d$
Subtract equation 2 to equation 1 to obtain:
$\begin{array}{cccc}
& &118 &= &a +11d
\\&-&&&&&
\\& &146 &= &a + 7d
\\&&\text{_____}&\text{______}&\text{_________}
\\&&-28 &= &4d\end{array}$
Divide both sides by $4$ to obtain:
$\dfrac{-28}{4}=\dfrac{4d}{4}
\\-7 = d$
Solve for $a$ by substituting $d=-7$ to Equation 2 to obtain:
$\require{cancel}
146 = a + 7d
\\146 = a + 7(-7)
\\146 = a + (-49)
\\146=a-49
\\146+49=a
\\195=a$
Thus, the first term is $a=195$.
The $n^{th}$ term $a_n$ of an arithmetic sequence is given by the formula:
$a_n=a+d(n-1)$
where
$a$ = first term
$d$ = common difference
The given sequence has $a= 195$ and $d=-7$.
Substituting these to the nth term formula gives:
$a_n = 195 + (-7)(n-1)
\\a_n=195-7(n-1)$
