1. The problem is to find the formula for the geometric sequence given points (1, 9) and (2, \frac{27}{5}). 2. Recall the formula for the $n$-th term of a geometric sequence: $$a_n = a_1 \cdot r^{n-1}$$ where $a_1$ is the first term and $r$ is the common ratio. 3. From the point $(1,9)$, we know: $$a_1 = 9$$ 4. Using the point $(2, \frac{27}{5})$, substitute $n=2$: $$a_2 = a_1 \cdot r^{2-1} = 9r = \frac{27}{5}$$ 5. Solve for $r$: $$9r = \frac{27}{5}$$ $$r = \frac{27}{5} \div 9 = \frac{27}{5} \cdot \frac{1}{9} = \frac{27}{45} = \frac{3}{5}$$ 6. Therefore, the formula for the sequence is: $$a_n = 9 \cdot \left(\frac{3}{5}\right)^{n-1}$$ 7. This formula means each term is obtained by multiplying the previous term by $\frac{3}{5}$, starting from 9. Final answer: $$a_n = 9 \left(\frac{3}{5}\right)^{n-1}$$