To find the 9th term of the sequence, we first observe the pattern in the given terms:
- The first term is \( a_1 = 15 \).
- The second term is \( a_2 = 10 \).
- The third term is \( a_3 = \frac{20}{3} \).
We need to see if there is a discernible pattern or formula that dictates how this sequence progresses.
Calculating the third term: \[ a_3 = \frac{20}{3} \approx 6.67 \]
Next, let's calculate the differences between consecutive terms:
- \( a_2 - a_1 = 10 - 15 = -5 \)
- \( a_3 - a_2 = \frac{20}{3} - 10 = \frac{20}{3} - \frac{30}{3} = -\frac{10}{3} \approx -3.33 \)
These terms decrease, but not in a consistent arithmetic or geometric fashion. Let's check the ratio of consecutive terms as well, just in case:
- The ratio \( \frac{a_2}{a_1} = \frac{10}{15} = \frac{2}{3} \)
- The ratio \( \frac{a_3}{a_2} = \frac{\frac{20}{3}}{10} = \frac{20}{30} = \frac{2}{3} \)
It seems that from the second term onward, each term is \( \frac{2}{3} \) of the previous term:
Assuming this geometric trend continues, the \( n \)-th term can be characterized by the following formula for \( n \geq 2 \):
\[ a_n = a_{n-1} \cdot \frac{2}{3} \] So we can calculate further terms: \[ \begin{align*} a_4 &= a_3 \cdot \frac{2}{3} = \frac{20}{3} \cdot \frac{2}{3} = \frac{40}{9} \approx 4.44 \ a_5 &= a_4 \cdot \frac{2}{3} = \frac{40}{9} \cdot \frac{2}{3} = \frac{80}{27} \approx 2.963 \ a_6 &= a_5 \cdot \frac{2}{3} = \frac{80}{27} \cdot \frac{2}{3} = \frac{160}{81} \approx 1.975 \ a_7 &= a_6 \cdot \frac{2}{3} = \frac{160}{81} \cdot \frac{2}{3} = \frac{320}{243} \approx 1.316 \ a_8 &= a_7 \cdot \frac{2}{3} = \frac{320}{243} \cdot \frac{2}{3} = \frac{640}{729} \approx 0.876 \ a_9 &= a_8 \cdot \frac{2}{3} = \frac{640}{729} \cdot \frac{2}{3} = \frac{1280}{2187} \end{align*} \]
Now, let's calculate \( \frac{1280}{2187} \) and round to the nearest thousandth:
\[ \frac{1280}{2187} \approx 0.585 \]
Thus, the 9th term rounded to the nearest thousandth is:
\[ \boxed{0.585} \]
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