In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio, known as the common ratio.

Question

To find the common ratio for the given sequence:

1. The first term \( a_1 \) is \( -\frac{1}{9} \).
2. The second term \( a_2 \) is \( -\frac{1}{3} \).
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:

\[
\begin{cases}
a_1 = -\frac{1}{9} \\
a_n = 3 \cdot a_{n-1} \quad \text{for } n \geq 2
\end{cases}
\]

\make this like a 9th grader wrote this

Answer 1

A geometric sequence is a special kind of sequence where each term is found by multiplying the previous term by the same number, called the common ratio.

Let's figure out the common ratio for this specific sequence:

1. The first term \( a_1 \) is \( -\frac{1}{9} \).
2. The second term \( a_2 \) is \( -\frac{1}{3} \).

To find the common ratio \( r \), we divide the second term by the first term:

\[ r = \frac{a_2}{a_1} = \frac{-\frac{1}{3}}{-\frac{1}{9}} = \frac{1/3}{1/9} = \frac{1}{3} \times \frac{9}{1} = 3 \]

This tells us that the common ratio \( r \) is 3.

Now, let's check that this ratio works for the other terms in the sequence:

• For the third term: \[ \frac{a_3}{a_2} = \frac{-1}{-\frac{1}{3}} = 3 \]

• For the fourth term: \[ \frac{a_4}{a_3} = \frac{-3}{-1} = 3 \]

• For the fifth term: \[ \frac{a_5}{a_4} = \frac{-9}{-3} = 3 \]

Since all the ratios are equal to 3, we can be sure that the common ratio is indeed 3.

Next, we can write a recursive formula for the geometric sequence. The formula looks like this:

\[ a_n = r \cdot a_{n-1} \]

For our specific sequence, this turns into:

\[ a_n = 3 \cdot a_{n-1} \quad \text{for } n \geq 2 \]

And we need to state the first term as well:

\[ a_1 = -\frac{1}{9} \]

So the final recursive formula for the sequence is:

\[ \begin{cases} a_1 = -\frac{1}{9} \ a_n = 3 \cdot a_{n-1} \quad \text{for } n \geq 2 \end{cases} \]

That’s it! We've figured out the common ratio and how to write the sequence in a mathematical way!