The common ratio of geometric progression {an} is q, and the product of the first n terms is t n, which satisfies the conditions A 1 > 1, A9A 100- 1 > 0.

(a99- 1)/(a 100- 1)< 0。

Give the following conclusions:

①0 < q < 1；

②a99？ a 10 1- 1 < 0；

③ The value of ③T 100 in TN is the largest;

④ Make the maximum natural number n of TN > 1 equal to 198.

The correct conclusion is ()

A.①②④ B.②④ C.①② D.①②③④

∵a99a 100- 1>0,∴a 12？ q 197> 1,∴(a 1？ q98)2> 1。

∵a 1> 1,∴q>0.

∫(a99- 1)/(a 100- 1)< 0，∴ A99 > 1，A 100 < 1。

∵a99？ a 10 1=a 100^2； 0