https://file.aiquickdraw.com/mathbot/file/1730342569668enzz7l3z.jpg

Let's solve this step-by-step.

### Step 1: Analyze the Series
The series given is:
\[
9 - 11 + \frac{121}{9} - \frac{1331}{81} + \dots
\]
The third term (\(n = 3\)) is \(\frac{121}{9}\).

### Step 2: Form of the Series and Common Ratio
To understand the structure of the series, let's observe the terms:
1. The first term is \(9\).
2. The second term is \(-11\).
3. The third term is \(\frac{121}{9}\).
4. The fourth term is \(-\frac{1331}{81}\).

We can see that each term alternates in sign, and there seems to be a common ratio that involves powers of \(\frac{11}{9}\).

Let's verify if this series follows a pattern \(a, ar, ar^2, \ldots\), where:
- \(a = 9\)
- \(r = -\frac{11}{9}\)

To check:
- The second term would be \(9 \cdot \left(-\frac{11}{9}\right) = -11\), which matches.
- The third term would be \(9 \cdot \left(-\frac{11}{9}\right)^2 = 9 \cdot \frac{121}{81} = \frac{121}{9}\), which matches.
- The fourth term would be \(9 \cdot \left(-\frac{11}{9}\right)^3 = -\frac{1331}{81}\), which matches.

Thus, the common ratio \(r\) is \(-\frac{11}{9}\).

### Step 3: Solving for \(a\)
The question states:
\[
ar^2 = a \left(-\frac{11}{9}\right)^2
\]
and asks us to find the value of \(a\).

Since we already identified \(a = 9\) from the series, we substitute this value into the expression.

### Final Answer
The value of \(a\) is \(9\).

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Would you like more details, or do you have any questions?

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**Follow-up Questions:**
1. How can we determine the general term in a geometric series?
2. What would the sum of the first few terms of this series be?
3. How does the alternating sign affect the convergence of the series?
4. Can we generalize this pattern to find other terms?
5. How does the common ratio influence the growth of the series?

**Tip:** In a geometric series, the common ratio is the key to finding each subsequent term—just multiply the previous term by the ratio.

In the series 9 - 11 + 121/9 - 1331/81 + ⋯, the n = 3 term is 121/9. If this is to equal ar^2 = a(-11/9)^2, then a = ___ .

Understand how to solve for 'a' in a geometric series problem with alternating signs. The problem involves identifying the common ratio and calculating terms of the series, ideal for high school students in Grades 10-12.

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