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 .
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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 \( (-2, 4) \).