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.
<span>From F.S 2, let the series be </span><span><span>a,b,ab,a<span>b2</span>,<span>a2</span><span>b3</span>,<span>a3</span><span>b5</span></span><span>a,b,ab,a<span>b2</span>,<span>a2</span><span>b3</span>,<span>a3</span><span>b5</span></span></span><span>. Now we know that </span><span><span>a<span>b2</span>=1</span><span>a<span>b2</span>=1</span></span>. The required difference =<span><span><span>a3</span><span>b5</span>−ab=ab(<span>a2</span><span>b4</span>−1)=ab[(a<span>b2</span><span>)2</span>−1]</span><span><span>a3</span><span>b5</span>−ab=ab(<span>a2</span><span>b4</span>−1)=ab[(a<span>b2</span><span>)2</span>−1]</span></span><span>= 0.Sufficient.</span>
1. The sequence shown in the problem is a Geometric sequence, because it has a common ratio. This means that when you multiply one of those numbers by the common ratio, you obtain the next number.
2. Then, you can calculate the common ratio as following: