18) In the following list, only one triplet of numbers cannot represent terms of a geometric progression (the terms do not necessarily have to be consecutive). Which is this triplet?
(A) $3 ; 6 ; 96$
(B) $3 ; 1 ; \frac{1}{81}$
(C) $3 ; 3.3 ; 3.993$
(D) $3 ; 6 ; 48$
(E) $3 ; 5 ; 9$.
18) The answer is (E). Indicating with $q$ the ratio of the progression, it is observed that in (A) and (D) the numbers can belong to a progression with $q=2$ and the first term $3(3 \cdot 2=6$, $3 \cdot 2^{4}=48,3 \cdot 2^{5}=96$ ). The numbers in (B) can belong to a progression with $q=\frac{1}{3}$ and the first term $3\left(3 \cdot \frac{1}{3}=1,3 \cdot\left(\frac{1}{3}\right)^{5}=\frac{1}{81}\right)$. The numbers in (C) can belong to a progression with $q=1.1$ and the first term $3\left(3 \cdot 1.1=3.3,3 \cdot(1.1)^{3}=3.993\right)$. On the other hand, the three terms in (E) cannot belong to the same geometric progression, otherwise one would have $\left(\frac{3}{5}\right)^{k}=\left(\frac{5}{9}\right)^{h}$, with $h$ and $k$ integers, which has no solutions.