13. The sum to infinity for a geometric series is \( \frac{3}{2} \) and the sum of the first three terms is \( \frac{14}{9} \). Find the first three terms of this series. 14. For which values of \( x \) will the series \( 1+\frac{x}{2 x-1}+\left(\frac{x}{2 x-1}\right)^{2}+\left(\frac{x}{2 x-1}\right)^{3}+\ldots \) converge? 15. Express 0,342 as a common fraction.

Let's address each of your problems one by one. --- ### **Problem 13** **Given:** - The sum to infinity of a geometric series is \( \frac{3}{2} \). - The sum of the first three terms is \( \frac{14}{9} \). **Find:** The first three terms of the series. **Solution:** 1. **Sum to Infinity Formula:** \[ S_\infty = \frac{a}{1 - r} = \frac{3}{2} \] Solving for \( a \): \[ a = \frac{3}{2} (1 - r) \] 2. **Sum of the First Three Terms:** \[ S_3 = a + ar + ar^2 = \frac{14}{9} \] Substitute \( a = \frac{3}{2} (1 - r) \) into the equation: \[ \frac{3}{2} (1 - r)(1 + r + r^2) = \frac{14}{9} \] 3. **Simplify the Equation:** \[ (1 - r)(1 + r + r^2) = 1 - r^3 \] So, \[ \frac{3}{2} (1 - r^3) = \frac{14}{9} \] \[ 1 - r^3 = \frac{28}{27} \] \[ r^3 = 1 - \frac{28}{27} = -\frac{1}{27} \] \[ r = -\frac{1}{3} \] 4. **Find \( a \):** \[ a = \frac{3}{2} \left(1 - \left(-\frac{1}{3}\right)\right) = \frac{3}{2} \times \frac{4}{3} = 2 \] 5. **First Three Terms:** \[ \begin{align*} a &= 2 \\ ar &= 2 \times \left(-\frac{1}{3}\right) = -\frac{2}{3} \\ ar^2 &= 2 \times \left(\frac{1}{9}\right) = \frac{2}{9} \end{align*} \] **Answer:** The first three terms are \( 2 \), \( -\frac{2}{3} \), and \( \frac{2}{9} \). --- ### **Problem 14** **Given:** The series: \[ 1 + \frac{x}{2x - 1} + \left(\frac{x}{2x - 1}\right)^2 + \left(\frac{x}{2x - 1}\right)^3 + \ldots \] **Find:** The values of \( x \) for which the series converges. **Solution:** This is an **infinite geometric series** with: - First term (\( a \)) = 1 - Common ratio (\( r \)) = \( \frac{x}{2x - 1} \) **Convergence Criterion:** \[ |r| < 1 \Rightarrow \left| \frac{x}{2x - 1} \right| < 1 \] **Solving the Inequality:** 1. \[ \left| \frac{x}{2x - 1} \right| < 1 \Rightarrow |x| < |2x - 1| \] 2. **Case 1:** \( 2x - 1 > 0 \) (i.e., \( x > \frac{1}{2} \)) \[ x < 2x - 1 \Rightarrow x > 1 \] 3. **Case 2:** \( 2x - 1 < 0 \) (i.e., \( x < \frac{1}{2} \)) - **Subcase a:** \( x \geq 0 \) \[ x < -2x + 1 \Rightarrow 3x < 1 \Rightarrow x < \frac{1}{3} \] - **Subcase b:** \( x < 0 \) \[ -x < -2x + 1 \Rightarrow x < 1 \quad (\text{Always true for } x < 0) \] **Combined Solutions:** - From **Case 1:** \( x > 1 \) - From **Case 2:** \( x < \frac{1}{3} \) **Answer:** The series converges for all real numbers \( x \) with \( x < \frac{1}{3} \) or \( x > 1 \). --- ### **Problem 15** **Given:** The decimal \( 0.342 \) **Find:** Its representation as a common fraction. **Solution:** 1. Express the decimal as a fraction: \[ 0.342 = \frac{342}{1000} \] 2. **Simplify the Fraction:** - **Find the Greatest Common Divisor (GCD) of 342 and 1000:** - \( 342 = 2 \times 3^2 \times 19 \) - \( 1000 = 2^3 \times 5^3 \) - **GCD:** 2 - Divide numerator and denominator by GCD: \[ \frac{342}{1000} = \frac{342 \div 2}{1000 \div 2} = \frac{171}{500} \] **Answer:** \( 0.342 \) expressed as a common fraction is \( \frac{171}{500} \).