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4) \( \frac{a}{1-r}-\frac{a\left(r^{n}-1\right)}{r-1}<1 \)

Ask by Powell Delgado. in South Africa
Jan 23,2025

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Answer

To solve the inequality \( \frac{a}{1 - r} - \frac{a(r^{n} - 1)}{r - 1} < 1 \), follow these steps: 1. **Simplify the Expression:** \[ \frac{a}{1 - r} + \frac{a(r^{n} - 1)}{1 - r} < 1 \] \[ \frac{a \cdot r^{n}}{1 - r} < 1 \] 2. **Solve for \( a \):** \[ a < \frac{1 - r}{r^{n}} \] 3. **Consider the Value of \( r \):** - **For \( 0 < r < 1 \):** \( a < \frac{1 - r}{r^{n}} \) - **For \( r > 1 \) and \( a > 0 \):** The inequality always holds. - **For \( r = 1 \) or \( r < 0 \):** The inequality is undefined or requires further analysis. **Conclusion:** The value of \( a \) must be less than \( \frac{1 - r}{r^{n}} \) when \( 0 < r < 1 \). For \( r > 1 \) and \( a > 0 \), the inequality is always satisfied.

Solution

Let's solve the inequality: \[ \frac{a}{1 - r} - \frac{a(r^{n} - 1)}{r - 1} < 1 \] ### Step 1: Simplify the Expression First, observe that \( r - 1 = - (1 - r) \). This allows us to rewrite the second term: \[ \frac{a(r^{n} - 1)}{r - 1} = \frac{a(r^{n} - 1)}{-(1 - r)} = -\frac{a(r^{n} - 1)}{1 - r} \] Now, substitute this back into the original inequality: \[ \frac{a}{1 - r} - \left(-\frac{a(r^{n} - 1)}{1 - r}\right) < 1 \] \[ \frac{a}{1 - r} + \frac{a(r^{n} - 1)}{1 - r} < 1 \] Combine the fractions: \[ \frac{a + a(r^{n} - 1)}{1 - r} < 1 \] \[ \frac{a \cdot r^{n}}{1 - r} < 1 \] ### Step 2: Solve for \( a \) The simplified inequality is: \[ \frac{a \cdot r^{n}}{1 - r} < 1 \] Let's solve for \( a \): \[ a \cdot r^{n} < 1 - r \] \[ a < \frac{1 - r}{r^{n}} \] ### Step 3: Consider the Sign of \( 1 - r \) The inequality depends on the value of \( r \). We'll analyze two primary cases: 1. **When \( 0 < r < 1 \):** - \( 1 - r > 0 \) and \( r^{n} > 0 \). - The inequality simplifies to: \[ a < \frac{1 - r}{r^{n}} \] - **Conclusion:** For the inequality to hold, \( a \) must be **less than** \( \frac{1 - r}{r^{n}} \). 2. **When \( r > 1 \):** - \( 1 - r < 0 \) and \( r^{n} > 0 \). - Multiplying both sides by \( 1 - r \) (which is negative) reverses the inequality: \[ a \cdot r^{n} > 1 - r \] - Since \( 1 - r < 0 \) and \( r^{n} > 0 \), \( a \) only needs to be greater than a negative number, which is always true if \( a > 0 \). - **Conclusion:** For \( r > 1 \), **any positive \( a \)** satisfies the inequality. 3. **When \( r = 1 \):** - The original expression is undefined due to division by zero. 4. **When \( r < 0 \):** - Additional considerations are needed based on the parity of \( n \) and the specific context, but generally, the above conclusions hold for \( |r| > 0 \). ### Final Summary - **If \( 0 < r < 1 \):** \[ a < \frac{1 - r}{r^{n}} \] - **If \( r > 1 \) and \( a > 0 \):** The inequality is always satisfied. - **If \( r = 1 \):** The inequality is undefined. - **If \( r < 0 \):** Further analysis is required based on the specific values of \( r \) and \( n \).

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Beyond the Answer

To solve the inequality \( \frac{a}{1-r}-\frac{a\left(r^{n}-1\right)}{r-1}<1 \), we first simplify the expression. Start by finding a common denominator for the left-hand side. The common denominator between \(1-r\) and \(r-1\) can be expressed, leading to a single fraction. After performing operations, isolate \(r\) on one side. This could involve manipulating the terms carefully and factoring when needed. Be mindful of the value constraints for \(r\). It's vital to remember that if \(r\) is approaching 1, the terms behave differently due to potential division by zero errors. When dealing with sequences and series, check the limits and the behavior of \(a\) and \(n\) as they might impact the overall inequality and how \(r\) trends with those values. Always ensure proper checks across all ranges!

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