A6. The first term of a geometric sequence with only positive terms is 27, and the seventh term is $\frac{1}{27}$. The twelfth term is equal to:
(A) $\frac{1}{3^{8}}$
(B) $\pm \frac{1}{3^{9}}$
(C) $-\frac{1}{6561}$
(D) 1
(E) 10

## II. PART
A6. Considering that $a_{7}=a_{1} \cdot q^{6}$, we have $\frac{1}{27}=27 \cdot q^{6}$, from which we get $q= \pm \frac{1}{3}$. Since the sequence has only positive terms, the twelfth term is $a_{12}=a_{1} \cdot q^{11}=27 \cdot\left(\frac{1}{3}\right)^{11} \stackrel{=}{=}$ $\left(\frac{1}{3}\right)^{8}=\frac{1}{3^{8}}$.

## II. PART