find the sixth term of a geometric sequence for which a2=228 and r= 1/4

an=a1(r)^(n-1) is geometric a1=first term r=common ratio given a2=228 r=1/4 a2=a1(1/4)^(2-1)=228 a1(1/4)^1=228 a1(1/4)=228 times both side sby 4/1 a1=912 an=912(1/4)^(n-1) 6th term a6 a6=912(1/4)^(6-1) a6=912(1/4)^5) a6=912(1/1024) a6=57/64 6th term is 57/64

57/64 is the answer hope I could help smiley