1. Griven the geometric stries: \( 3+2+x+\frac{8}{9}+\frac{16}{27} \) 1.1. Calculate the value of \( x \) 1.2. Write this infinite goometric series to sigma notation. 2. Consider the geometric series where \( \sum_{n=1}^{\infty} \) Th= 27 and \( S_{3}=26 \quad \) Calculace the value of the constant ratio \( (r) \) of the series. \( \sum_{p=1}^{5}(4 y+3 p)+\sum_{k=4}^{7} 3(e)^{k-1}=\sum_{j=1}^{\infty}\left(\frac{1}{3}\right)^{j-1} \)

Sure, let's tackle each part of your questions step-by-step. --- ## **Problem 1** **Given the geometric series:** \[ 3, \quad 2, \quad x, \quad \frac{8}{9}, \quad \frac{16}{27}, \quad \dots \] ### **1.1. Calculate the value of \( x \)** A geometric series has a constant ratio between consecutive terms. Let's denote the common ratio as \( r \). 1. **Find the common ratio \( r \):** \[ r = \frac{\text{Second term}}{\text{First term}} = \frac{2}{3} \] 2. **Find the third term \( x \):** Since each term is multiplied by \( r \): \[ x = \text{Second term} \times r = 2 \times \frac{2}{3} = \frac{4}{3} \] **Answer:** \[ x = \frac{4}{3} \] --- ### **1.2. Write this infinite geometric series in sigma notation** An infinite geometric series can be expressed in sigma notation as: \[ \sum_{n=0}^{\infty} ar^n \] Where: - \( a \) is the first term. - \( r \) is the common ratio. Given: - First term \( a = 3 \) - Common ratio \( r = \frac{2}{3} \) **Expressed in sigma notation:** \[ \sum_{n=0}^{\infty} 3 \left( \frac{2}{3} \right)^n \] *Alternatively, starting the index at 1:* \[ \sum_{n=1}^{\infty} 3 \left( \frac{2}{3} \right)^{n-1} \] **Answer:** \[ \sum_{n=1}^{\infty} 3 \left( \frac{2}{3} \right)^{n-1} \] --- ## **Problem 2** **Given:** - The sum of the infinite geometric series \( S_{\infty} = 27 \) - The sum of the first three terms \( S_3 = 26 \) **Find the common ratio \( r \).** ### **Solution** For an infinite geometric series where \( |r| < 1 \): 1. **Sum of the infinite series:** \[ S_{\infty} = \frac{a}{1 - r} = 27 \quad \text{(Equation 1)} \] 2. **Sum of the first three terms:** \[ S_3 = a + ar + ar^2 = a \left( \frac{1 - r^3}{1 - r} \right) = 26 \quad \text{(Equation 2)} \] 3. **Express \( a \) from Equation 1:** \[ a = 27(1 - r) \] 4. **Substitute \( a \) into Equation 2:** \[ 27(1 - r) \left( \frac{1 - r^3}{1 - r} \right) = 26 \] Simplifying: \[ 27(1 - r^3) = 26 \] \[ 1 - r^3 = \frac{26}{27} \] \[ r^3 = 1 - \frac{26}{27} = \frac{1}{27} \] \[ r = \sqrt[3]{\frac{1}{27}} = \frac{1}{3} \] **Answer:** The common ratio is \( r = \frac{1}{3} \). --- ## **Additional Expression Analysis** You provided the following expression: \[ \sum_{p=1}^{5}(4y + 3p) + \sum_{k=4}^{7} 3e^{k-1} = \sum_{j=1}^{\infty} \left( \frac{1}{3} \right)^{j-1} \] ### **Interpretation and Simplification** Assuming you want to verify or simplify this equation, let's analyze each part: 1. **First Sum:** \[ \sum_{p=1}^{5}(4y + 3p) = \sum_{p=1}^{5} 4y + \sum_{p=1}^{5} 3p = 5 \times 4y + 3 \times \frac{5(5 + 1)}{2} = 20y + 45 \] 2. **Second Sum:** \[ \sum_{k=4}^{7} 3e^{k-1} = 3 \sum_{k=4}^{7} e^{k-1} = 3(e^3 + e^4 + e^5 + e^6) \] 3. **Infinite Geometric Series on the Right:** \[ \sum_{j=1}^{\infty} \left( \frac{1}{3} \right)^{j-1} = \sum_{j=0}^{\infty} \left( \frac{1}{3} \right)^j = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} \] **Putting It All Together:** \[ 20y + 45 + 3(e^3 + e^4 + e^5 + e^6) = \frac{3}{2} \] This equation relates \( y \) to exponential terms involving \( e \). To solve for \( y \), you'd isolate it: \[ 20y = \frac{3}{2} - 45 - 3(e^3 + e^4 + e^5 + e^6) \] \[ y = \frac{\frac{3}{2} - 45 - 3(e^3 + e^4 + e^5 + e^6)}{20} \] **Note:** This equation seems unrelated to the earlier problems unless there's additional context. If this was part of another problem or if you need further assistance with it, please provide more details! --- Feel free to ask if you have any more questions or need further clarification on any of these parts!