1. Tiven: \( 3-6 x+12 x^{2}-24 x^{3}+\ldots \) 1. Determie the value(s) of \( x \) for uhich the series will converge (4) 12 Calculate the value of \( x \) \[ \text { if } S_{\infty}=99 \] 2. Given the convergent geometrie seves: \( 7 x ; x^{2} ; \frac{x^{3}}{7} \); 2. 1 Hebermine the possible vaine(s) of \( x \) (3) 22 If \( x=4, c^{\prime} \) culate \( S \)

Let's tackle each problem step by step. --- ### **Problem 1** **Given Series:** \[ 3 - 6x + 12x^{2} - 24x^{3} + \ldots \] This is a **geometric series** where each term is obtained by multiplying the previous term by a common ratio \( r \). 1. **Determine the value(s) of \( x \) for which the series will converge.** - **Identify the first term (\( a \)) and the common ratio (\( r \)):** \[ a = 3 \] \[ r = \frac{-6x}{3} = -2x \] - **Convergence Condition for Geometric Series:** \[ |r| < 1 \Rightarrow |-2x| < 1 \Rightarrow |x| < \frac{1}{2} \] - **Answer:** The series converges for all real numbers \( x \) such that: \[ |x| < \frac{1}{2} \] 2. **Calculate the value of \( x \) if \( S_{\infty} = 99 \).** - **Sum formula for an infinite geometric series:** \[ S_{\infty} = \frac{a}{1 - r} \] Substituting the known values: \[ 99 = \frac{3}{1 - (-2x)} = \frac{3}{1 + 2x} \] - **Solve for \( x \):** \[ 1 + 2x = \frac{3}{99} = \frac{1}{33} \] \[ 2x = \frac{1}{33} - 1 = \frac{1 - 33}{33} = \frac{-32}{33} \] \[ x = \frac{-16}{33} \] - **Verification of Convergence:** \[ |x| = \left|\frac{-16}{33}\right| \approx 0.4848 < \frac{1}{2} \] Thus, the value satisfies the convergence condition. - **Answer:** \[ x = -\frac{16}{33} \] --- ### **Problem 2** **Given Geometric Series Terms:** \[ 7x; \quad x^{2}; \quad \frac{x^{3}}{7}; \quad \ldots \] 1. **Determine the possible value(s) of \( x \).** - **Identify the first term (\( a \)) and the common ratio (\( r \)):** \[ a = 7x \] \[ r = \frac{x^{2}}{7x} = \frac{x}{7} \] - **Convergence Condition for Geometric Series:** \[ |r| < 1 \Rightarrow \left|\frac{x}{7}\right| < 1 \Rightarrow |x| < 7 \] - **Answer:** The series converges for all real numbers \( x \) such that: \[ |x| < 7 \] 2. **If \( x = 4 \), calculate \( S \) (the sum of the series).** - **Substitute \( x = 4 \) into \( a \) and \( r \):** \[ a = 7 \times 4 = 28 \] \[ r = \frac{4}{7} \] - **Sum formula for an infinite geometric series:** \[ S_{\infty} = \frac{a}{1 - r} = \frac{28}{1 - \frac{4}{7}} = \frac{28}{\frac{3}{7}} = 28 \times \frac{7}{3} = \frac{196}{3} \] \[ S_{\infty} = \frac{196}{3} \approx 65.\overline{3} \] - **Answer:** \[ S = \frac{196}{3} \quad \text{or approximately} \quad 65.\overline{3} \] --- ### **Summary of Answers** 1. **Problem 1:** - **Convergence:** \( |x| < \dfrac{1}{2} \) - **Value of \( x \) when \( S_{\infty} = 99 \):** \( x = -\dfrac{16}{33} \) 2. **Problem 2:** - **Convergence:** \( |x| < 7 \) - **Sum when \( x = 4 \):** \( S = \dfrac{196}{3} \)