Answer
$$12,10,8,6,4$$
Work Step by Step
$$\eqalign{
& {\text{arithmetic;}}\,\,\,{a_2} = 10,\,\,\,d = - 2 \cr
& {\text{Substituting values for }}{a_2}{\text{ and }}d{\text{ in the formula }} \cr
& {a_n} = {a_1} + \left( {n - 1} \right)d \cr
& {a_2} = {a_1} + \left( {2 - 1} \right)d \cr
& 10 = {a_1} + \left( {2 - 1} \right)\left( { - 2} \right) \cr
& {\text{Solve for }}{a_1} \cr
& 10 = {a_1} - 2 \cr
& {a_1} = 12 \cr
& \cr
& {\text{The }}n{\text{th Term of an Arithmetic Sequence is}} \cr
& {a_n} = {a_1} + \left( {n - 1} \right)d \cr
& {a_n} = 12 + \left( {n - 1} \right)\left( { - 2} \right) \cr
& {a_n} = 12 - 2n + 2 \cr
& {a_n} = - 2n + 14 \cr
& \cr
& {\text{Find }}{a_3},\,\,{a_4},\,\,\,{a_5} \cr
& {a_3} = - 2\left( 3 \right) + 14 = 8 \cr
& {a_4} = - 2\left( 4 \right) + 14 = 6 \cr
& {a_5} = - 2\left( 5 \right) + 14 = 4 \cr
& \cr
& {\text{The terms are:}} \cr
& 12,10,8,6,4 \cr} $$
