4) If \( (4-x), 3 \) and \( \sqrt{2 x+3} \) are the first three terms of an arithmetic sequence: \( \begin{array}{ll}\text { a) Determine the value of } x & \text { b) Which term is equal to }-23 ? \\ \text { 5) If } q ; 2 q+2 ; 5 q+3 \text { form an arithmetic sequence: } \\ \begin{array}{ll}\text { a) Determine the value of } q & \text { b) Determine the value of the } 15^{\text {th }} \text { term. } \\ \text { 6) If } 2 p-1 ; p+1 ; p-1 \text { form a geometric sequence and } p \neq 0\end{array} \\ \begin{array}{ll}\text { a) What are the first three terms? }\end{array} \\ \text { 7) If } 8 ; 2 x ; 2 y \text { forms an arithmetic sequence and } 2 x ; 2 y ; 36 \text { forms a geometric sequence deterr } \\ \text { the values of } x \text { and } y \text {. }\end{array} \)

Let's solve each problem step by step. --- ### **Problem 4** **Given:** - Three terms of an arithmetic sequence: \( (4 - x) \), \( 3 \), and \( \sqrt{2x + 3} \). #### **a) Determine the value of \( x \).** **Solution:** In an arithmetic sequence, the difference between consecutive terms is constant. Therefore: \[ 3 - (4 - x) = \sqrt{2x + 3} - 3 \] Simplify the left side: \[ 3 - 4 + x = x - 1 \] Set the differences equal: \[ x - 1 = \sqrt{2x + 3} - 3 \] Rearrange the equation: \[ x + 2 = \sqrt{2x + 3} \] Square both sides to eliminate the square root: \[ (x + 2)^2 = 2x + 3 \\ x^2 + 4x + 4 = 2x + 3 \] Bring all terms to one side: \[ x^2 + 2x + 1 = 0 \\ (x + 1)^2 = 0 \] **Solution:** \[ x = -1 \] **Verification:** - First term: \( 4 - (-1) = 5 \) - Second term: \( 3 \) - Third term: \( \sqrt{2(-1) + 3} = \sqrt{1} = 1 \) Check the common difference: \[ 3 - 5 = -2 \\ 1 - 3 = -2 \] The differences are equal, confirming \( x = -1 \) is correct. #### **b) Which term is equal to \( -23 \)?** **Solution:** First term (\( a_1 \)) = 5 Common difference (\( d \)) = \( 3 - 5 = -2 \) The \( n \)-th term of an arithmetic sequence is given by: \[ a_n = a_1 + (n - 1)d \] Set \( a_n = -23 \): \[ -23 = 5 + (n - 1)(-2) \\ -23 = 5 - 2n + 2 \\ -23 = 7 - 2n \\ -30 = -2n \\ n = 15 \] **Answer:** The 15th term is equal to \( -23 \). --- ### **Problem 5** **Given:** - Three terms of an arithmetic sequence: \( q \), \( 2q + 2 \), \( 5q + 3 \). #### **a) Determine the value of \( q \).** **Solution:** In an arithmetic sequence: \[ (2q + 2) - q = (5q + 3) - (2q + 2) \] Simplify: \[ q + 2 = 3q + 1 \\ q + 2 - 3q -1 = 0 \\ -2q +1 = 0 \\ 2q = 1 \\ q = \frac{1}{2} \] #### **b) Determine the value of the 15th term.** **Solution:** First term (\( a_1 \)) = \( q = \frac{1}{2} \) Common difference (\( d \)) = \( (2q + 2) - q = \frac{1}{2} + 2 = \frac{5}{2} \) The \( n \)-th term: \[ a_n = a_1 + (n - 1)d \] For \( n = 15 \): \[ a_{15} = \frac{1}{2} + 14 \times \frac{5}{2} = \frac{1}{2} + 35 = \frac{71}{2} = 35.5 \] **Answer:** The 15th term is \( \frac{71}{2} \) or \( 35.5 \). --- ### **Problem 6** **Given:** - Three terms form a geometric sequence: \( 2p - 1 \), \( p + 1 \), \( p - 1 \), with \( p \neq 0 \). #### **a) What are the first three terms?** **Solution:** For a geometric sequence, the ratio between consecutive terms is constant: \[ \frac{p + 1}{2p - 1} = \frac{p - 1}{p + 1} \] Cross-multiply: \[ (p + 1)^2 = (2p - 1)(p - 1) \] Expand both sides: \[ p^2 + 2p + 1 = 2p^2 - 3p + 1 \] Bring all terms to one side: \[ p^2 + 2p + 1 - 2p^2 + 3p - 1 = 0 \\ -p^2 + 5p = 0 \\ p(p - 5) = 0 \] Since \( p \neq 0 \), \( p = 5 \). **First three terms:** \[ 2p - 1 = 9, \quad p + 1 = 6, \quad p - 1 = 4 \] **Answer:** The first three terms are \( 9 \), \( 6 \), and \( 4 \). --- ### **Problem 7** **Given:** - Two sequences: - Arithmetic: \( 8 \), \( 2x \), \( 2y \) - Geometric: \( 2x \), \( 2y \), \( 36 \) **Determine \( x \) and \( y \).** **Solution:** **Step 1: Arithmetic Sequence Condition** For arithmetic sequence \( 8 \), \( 2x \), \( 2y \): \[ 2x - 8 = 2y - 2x \\ 4x - 8 = 2y \\ 2x - 4 = y \quad \text{(Equation 1)} \] **Step 2: Geometric Sequence Condition** For geometric sequence \( 2x \), \( 2y \), \( 36 \): \[ \frac{2y}{2x} = \frac{36}{2y} \\ \frac{y}{x} = \frac{36}{2y} \\ y^2 = 18x \quad \text{(Equation 2)} \] **Step 3: Substitute Equation 1 into Equation 2** From Equation 1: \[ y = 2x - 4 \] Substitute into Equation 2: \[ (2x - 4)^2 = 18x \\ 4x^2 - 16x + 16 = 18x \\ 4x^2 - 34x + 16 = 0 \] **Step 4: Solve the Quadratic Equation** \[ x = \frac{34 \pm \sqrt{34^2 - 4 \cdot 4 \cdot 16}}{2 \cdot 4} \\ x = \frac{34 \pm \sqrt{1156 - 256}}{8} \\ x = \frac{34 \pm \sqrt{900}}{8} \\ x = \frac{34 \pm 30}{8} \] **Possible Values:** 1. \( x = \frac{34 + 30}{8} = \frac{64}{8} = 8 \) - \( y = 2(8) - 4 = 12 \) 2. \( x = \frac{34 - 30}{8} = \frac{4}{8} = 0.5 \) - \( y = 2(0.5) - 4 = -3 \) **Verification:** - **Case 1:** \( x = 8 \), \( y = 12 \) - Arithmetic sequence: \( 8 \), \( 16 \), \( 24 \) - Common difference: \( 8 \) - Geometric sequence: \( 16 \), \( 24 \), \( 36 \) - Common ratio: \( \frac{24}{16} = \frac{36}{24} = 1.5 \) - **Case 2:** \( x = 0.5 \), \( y = -3 \) - Arithmetic sequence: \( 8 \), \( 1 \), \( -6 \) - Common difference: \( -7 \) - Geometric sequence: \( 1 \), \( -6 \), \( 36 \) - Common ratio: \( \frac{-6}{1} = \frac{36}{-6} = -6 \) Both cases satisfy the conditions. **Answer:** There are two possible solutions: - \( x = 8 \) and \( y = 12 \) - \( x = 0.5 \) and \( y = -3 \) ---