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>
