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 .

To determine the possible values of \( x \) for which the given sequence forms a convergent geometric sequence, let's analyze the sequence step by step. ### Given Sequence: \[ \frac{3}{(x-1)^2},\ \frac{1}{(x-1)},\ \frac{1}{3},\ \frac{(x-1)}{9},\ \ldots \] ### Step 1: Identify the Common Ratio (\( r \)) A geometric sequence has the form: \[ a, \ ar, \ ar^2, \ ar^3, \ \ldots \] where \( a \) is the first term and \( r \) is the common ratio. From the given sequence: - First term (\( a \)): \( \frac{3}{(x-1)^2} \) - Second term: \( \frac{1}{(x-1)} \) Using the relationship between the first two terms: \[ ar = \frac{1}{(x-1)} \] Substituting \( a \): \[ \frac{3}{(x-1)^2} \cdot r = \frac{1}{(x-1)} \] Solving for \( r \): \[ r = \frac{(x-1)}{3} \] ### Step 2: Verify Consistency with Subsequent Terms - Third term (\( ar^2 \)): \( \frac{1}{3} \) \[ \frac{3}{(x-1)^2} \cdot \left(\frac{(x-1)}{3}\right)^2 = \frac{1}{3} \] This simplifies correctly, verifying the consistency. - Fourth term (\( ar^3 \)): \( \frac{(x-1)}{9} \) \[ \frac{3}{(x-1)^2} \cdot \left(\frac{(x-1)}{3}\right)^3 = \frac{(x-1)}{9} \] This also simplifies correctly. ### Step 3: Determine Convergence A geometric sequence converges if the absolute value of the common ratio is less than 1: \[ |r| < 1 \] Substituting \( r \): \[ \left| \frac{(x-1)}{3} \right| < 1 \] \[ |x - 1| < 3 \] This leads to: \[ -3 < x - 1 < 3 \] \[ -2 < x < 4 \] ### Step 4: Consider Domain Restrictions The sequence has terms like \( \frac{1}{(x-1)} \) and \( \frac{3}{(x-1)^2} \), which are undefined when \( x = 1 \). Therefore, \( x = 1 \) must be excluded. ### **Conclusion:** The sequence converges for all real numbers \( x \) such that: \[ -2 < x < 4 \quad \text{and} \quad x \neq 1 \] **Final Answer:** All real numbers x between –2 and 4 except x = 1. In other words, –2 < x < 4 and x ≠ 1