From F.S 1, consider this series : 8, 1, 8, 8, 64, 64*8.
Again, consider the series 2, 1/4, 1/2, 1/8, 1/16, 1/(8*16). Clearly, the difference of the 6th and the 3rd term is different for them. Insufficient.
From F.S 2, let the series be a,b,ab,ab2,a2b3,a3b5a,b,ab,ab2,a2b3,a3b5. Now we know that ab2=1ab2=1. The required difference =a3b5−ab=ab(a2b4−1)=ab[(ab2)2−1]a3b5−ab=ab(a2b4−1)=ab[(ab2)2−1]= 0.Sufficient.