## Algebra 2 (1st Edition)

$a_n=\dfrac{111-13n}{5}$
We know that the general formula of an arithmetic sequence is given by $a_n= a_1+(n-1) d$ ...(1) Here, we have $a_7=a+7b=4$ ..(2) $a_{12}=a+12b=-9$ ..(3) Now, we will have to subtract equation (2) from (3). $5b =-13 \implies b=\dfrac{-13}{5}$ Equation (2) gives: $a_7=a+7(\dfrac{-13}{5})=4$ $a=\dfrac{111}{5}$ Thus, we find that the nth term is: $a_n=\dfrac{111-13n}{5}$