In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio, known as the common ratio.
To find the common ratio for the given sequence:
- The first term \( a_1 \) is \( -\frac{1}{9} \).
- The second term \( a_2 \) is \( -\frac{1}{3} \).
- To find the common ratio \( r \), divide the second term by the first term:
\[ r = \frac{a_2}{a_1} = \frac{-\frac{1}{3}}{-\frac{1}{9}} = \frac{1/3}{1/9} = \frac{1}{3} \cdot \frac{9}{1} = 3 \]
This calculation shows that the common ratio \( r \) is 3.
We can verify this ratio for the remaining terms:
\[ \frac{a_3}{a_2} = \frac{-1}{-\frac{1}{3}} = 3 \] \[ \frac{a_4}{a_3} = \frac{-3}{-1} = 3 \] \[ \frac{a_5}{a_4} = \frac{-9}{-3} = 3 \]
Since all ratios are equal to \( r = 3\), we confirm that the common ratio is indeed 3.
The recursive formula for a geometric sequence can be written as follows:
\[ a_n = r \cdot a_{n-1} \]
For this specific sequence, the recursive formula is:
\[ a_n = 3 \cdot a_{n-1} \quad \text{for } n \geq 2 \]
With the first term defined as:
\[ a_1 = -\frac{1}{9} \]
Thus, the complete recursive formula for the sequence is:
\[ \begin{cases} a_1 = -\frac{1}{9} \ a_n = 3 \cdot a_{n-1} \quad \text{for } n \geq 2 \end{cases} \]