bila suku ke-n deret geometri ditentukan oleh un =3/3n maka jumlah 6 suku pertama deret itu sama dengan......

U1 =a = 3/3(1) = 1
U2 = 3/3(2) = 3/6 = 1/2
u3 = 3/3(3) = 3/9 = 1/3
u4 = 3/3(4) = 3/12 = 1/4
U5 = 3/3(5) = 3/15 = 1/5
u6 = 3/3(6) = 3/18 = 1/6
S6 = 1 + 1/2+1/3+1/4+1/5+1/6
   = (60 + 30 + 20 + 15 +12 + 10)/60
=147 / 60
= 49/20