Given the arithmetic sequence with A3 =12 and A9=14 find A1and A30

Answer 1

(14-12)/(9-3)= 2/6 = 1/3 = d = common difference

A1 = A3-2d=12-2/3 = 11 1/3

A30=A1+d(30-1)

A30 =11 1/3 +(1/3)(29)=(34+29)/3=63/3=21

12,13,14,...20,21

are A3,A6,A9,...A27,A30

every 3rd term increases by 1