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 $$ S $$

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### **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:**  
     $$
     x = -\frac{16}{33}
     $$

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### **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:**  
     $$
     S = \frac{196}{3} \quad \text{or approximately} \quad 65.\overline{3}
     $$

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### **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 $$ x = 4 $$:** $$ S = \dfrac{196}{3} $$
**Problem 1:** - The series converges when $$ |x| < \frac{1}{2} $$. - When $$ S_{\infty} = 99 $$, $$ x = -\frac{16}{33} $$. **Problem 2:** - The series converges when $$ |x| < 7 $$. - When $$ x = 4 $$, $$ S = \frac{196}{3} $$ or approximately 65.33.

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