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3.3 The following sequence forms a convergent geometric sequence: \( \frac{3}{(x-1)^{2}}+\frac{1}{(x-1)}+\frac{1}{3}+\frac{(x-1)}{9}+\ldots \) 3.3.1 Determine the possible values of x .

Ask by Garza Gough. in South Africa
Jan 03,2025

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Answer

All real numbers \( x \) such that \( -2 < x < 4 \) and \( x \neq 1 \).

Solution

To determine the possible values of \( x \) for the given convergent geometric sequence: \[ \frac{3}{(x-1)^{2}} + \frac{1}{(x-1)} + \frac{1}{3} + \frac{(x-1)}{9} + \ldots \] we follow these steps: ### 1. Identify the Terms and Common Ratio A geometric sequence has terms in the form \( a, ar, ar^2, ar^3, \ldots \), where: - \( a \) is the first term. - \( r \) is the common ratio. For the given sequence: - **First Term (\( a \))**: \( \frac{3}{(x-1)^2} \) - **Second Term**: \( \frac{1}{(x-1)} \) - **Third Term**: \( \frac{1}{3} \) - **Fourth Term**: \( \frac{(x-1)}{9} \) ### 2. Determine the Common Ratio (\( r \)) The common ratio can be found by dividing the second term by the first term, the third term by the second term, and so on. These ratios should all be equal for the sequence to be geometric. \[ r = \frac{\text{Second Term}}{\text{First Term}} = \frac{\frac{1}{x-1}}{\frac{3}{(x-1)^2}} = \frac{(x-1)^2}{3(x-1)} = \frac{x-1}{3} \] Similarly, checking the ratio between the third and second terms: \[ r = \frac{\text{Third Term}}{\text{Second Term}} = \frac{\frac{1}{3}}{\frac{1}{x-1}} = \frac{(x-1)}{3} \] And between the fourth and third terms: \[ r = \frac{\text{Fourth Term}}{\text{Third Term}} = \frac{\frac{x-1}{9}}{\frac{1}{3}} = \frac{x-1}{3} \] All ratios confirm that \( r = \frac{x-1}{3} \). ### 3. Conditions for Convergence For a geometric sequence to be convergent, the absolute value of the common ratio must be less than 1: \[ \left| \frac{x-1}{3} \right| < 1 \] Solving this inequality: \[ -1 < \frac{x-1}{3} < 1 \\ -3 < x - 1 < 3 \\ -2 < x < 4 \] ### 4. Excluding Undefined Values Additionally, the terms involve denominators like \( (x-1) \) and \( (x-1)^2 \), which are undefined when \( x = 1 \). Therefore, \( x = 1 \) must be excluded from the solution. ### Final Answer All real numbers x between –2 and 4 except x = 1. In other words, –2 < x < 4 and x ≠ 1.

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To determine the possible values of \( x \) for the given geometric sequence \( \frac{3}{(x-1)^{2}}+\frac{1}{(x-1)}+\frac{1}{3}+\frac{(x-1)}{9}+\ldots \), we start by identifying the first term \( a = \frac{3}{(x-1)^2} \) and the common ratio \( r \). To find \( r \), we examine the ratio of the second term to the first term: \[ r = \frac{\frac{1}{(x-1)}}{\frac{3}{(x-1)^2}} = \frac{(x-1)}{3}. \] Similarly, we calculate the ratio between the third and second terms: \[ r = \frac{\frac{1}{3}}{\frac{1}{(x-1)}} = \frac{(x-1)}{3}. \] Lastly, the ratio between the fourth and third terms gives: \[ r = \frac{\frac{(x-1)}{9}}{\frac{1}{3}} = \frac{(x-1)}{3}. \] Now, for the series to converge, the condition \( |r| < 1 \) must hold: \[ \left|\frac{(x-1)}{3}\right| < 1. \] This leads us to the inequalities: \[ -1 < \frac{(x-1)}{3} < 1. \] Multiplying through by 3 gives: \[ -3 < x - 1 < 3. \] Finally, adding 1 to all parts: \[ -2 < x < 4. \] Thus, the possible values of \( x \) come out to be in the open interval \( (-2, 4) \).

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