Answer
$${a_1} = 27$$
Work Step by Step
$$\eqalign{
& {a_{12}} = 60,\,\,\,{a_{20}} = 84 \cr
& \cr
& {\text{We obtain }}{a_{20}}{\text{ by adding the common difference to }}{a_{12}}{\text{ eight times}} \cr
& {a_{20}} = {a_{12}} + 8d \cr
& {\text{Substituting }}{a_{20}}{\text{ and }}{a_{12}} \cr
& 84 = 60 + 8d \cr
& {\text{Solve for }}d \cr
& 24 = 8d \cr
& d = 3 \cr
& \cr
& {\text{The }}n{\text{th Term of an Arithmetic Sequence is given by}} \cr
& {a_n} = {a_1} + \left( {n - 1} \right)d \cr
& {\text{For }}n = 12 \cr
& {a_{12}} = {a_1} + \left( {12 - 1} \right)d \cr
& 60 = {a_1} + \left( {12 - 1} \right)\left( 3 \right) \cr
& 60 = {a_1} + 33 \cr
& {a_1} = 27 \cr} $$
