\[ \begin{array}{l} \left(\frac{y}{2}\right)^{n}=\left(\frac{\sum}{2}\right)^{n-1} \\ 8=n-1 \\ \therefore n=9 \end{array} \] CTIVTTIES/ASSESSMENTS Classwork/Homework 1. \( k+1 ; k-1 ; 2 k-5 ; \ldots \) are the first three terms of a geometric sequence, where \( k<0 \) : (a) Calculate the value of \( k \). (b) Determine the sequence. (c) Determine the general term of the seog (d) Find the 10 th term. (c) Which ferm of the sequence is -59649 ? 2. Determine an expression for the \( n^{\text {th }} \) term for a geometric sequence whose \( 3^{\text {th }} \) term is -20 and the \( 6^{\text {th }} \) term is 160 . 3. \( x-4: x+2 ; 3 x+1: \ldots \) are the first three terms of a geometric sequence. Determine the general term of the sequence in terms of \( x \).

Let's solve each part of the problem step by step. ### Problem 1: Geometric Sequence Given the first three terms of a geometric sequence: \( k+1, k-1, 2k-5 \) where \( k < 0 \). #### (a) Calculate the value of \( k \). In a geometric sequence, the ratio between consecutive terms is constant. Therefore, we can set up the following equation based on the ratios: \[ \frac{k-1}{k+1} = \frac{2k-5}{k-1} \] Cross-multiplying gives: \[ (k-1)^2 = (k+1)(2k-5) \] Expanding both sides: \[ k^2 - 2k + 1 = 2k^2 - 5k + 2k - 5 \] This simplifies to: \[ k^2 - 2k + 1 = 2k^2 - 3k - 5 \] Rearranging gives: \[ 0 = 2k^2 - 3k - 5 - k^2 + 2k - 1 \] \[ 0 = k^2 - k - 6 \] Factoring the quadratic: \[ 0 = (k - 3)(k + 2) \] Thus, \( k = 3 \) or \( k = -2 \). Since \( k < 0 \), we have: \[ k = -2 \] #### (b) Determine the sequence. Substituting \( k = -2 \) into the terms: 1. First term: \( k + 1 = -2 + 1 = -1 \) 2. Second term: \( k - 1 = -2 - 1 = -3 \) 3. Third term: \( 2k - 5 = 2(-2) - 5 = -4 - 5 = -9 \) Thus, the sequence is: \[ -1, -3, -9 \] #### (c) Determine the general term of the sequence. The common ratio \( r \) can be calculated as: \[ r = \frac{-3}{-1} = 3 \] The general term \( a_n \) of a geometric sequence is given by: \[ a_n = a_1 \cdot r^{n-1} \] Where \( a_1 = -1 \) and \( r = 3 \): \[ a_n = -1 \cdot 3^{n-1} \] #### (d) Find the 10th term. Substituting \( n = 10 \): \[ a_{10} = -1 \cdot 3^{10-1} = -3^9 \] Calculating \( 3^9 \): \[ 3^9 = 19683 \] Thus: \[ a_{10} = -19683 \] #### (e) Which term of the sequence is -59649? We need to find \( n \) such that: \[ -1 \cdot 3^{n-1} = -59649 \] This simplifies to: \[ 3^{n-1} = 59649 \] Calculating \( 3^{n-1} \): \[ 3^{n-1} = 3^{10} \quad \text{(since } 3^{10} = 59049 \text{)} \] Thus: \[ n - 1 = 10 \implies n = 11 \] ### Problem 2: Expression for the \( n^{\text{th}} \) term Given the \( 3^{\text{rd}} \) term is -20 and the \( 6^{\text{th}} \) term is 160. Let the first term be \( a \) and the common ratio be \( r \). The terms can be expressed as: \[ a r^2 = -20 \quad \text{(1)} \] \[ a r^5 = 160 \quad \text{(2)} \] Dividing equation (2) by equation (1): \[ \frac{a r^5}{a r^2} = \frac{160}{-20} \] This simplifies to: \[ r^3 = -8 \implies r = -2 \] Substituting \( r \) back into equation (1): \[ a (-2)^2 = -20 \implies 4a = -20 \implies a = -5 \] Thus, the general term is: \[ a_n = -5 \cdot (-2)^{n-1} \] ### Problem 3: General term of the sequence Given the first three terms: \( x-4, x+2, 3x+1 \). Setting up the ratio: \[ \frac{x+2}{x-4} = \frac{3x+1}{x+2} \] Cross-multiplying gives: \[ (x+2)^2 = (x-4)(3x+1) \] Expanding both sides: \[ x^2 + 4x + 4 = 3x^2 - 12x + x - 4 \] This simplifies to: \[ x^2 + 4x + 4 = 3x^2 - 11x - 4 \] Rearranging gives: \[ 0 = 2x^2 - 15x - 8 \] Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot (-8)}}{2 \cdot 2} \] Calculating the discriminant: \[ = \frac{15 \pm \sqrt{225 + 64}}{4} = \frac{15 \pm \sqrt{289}}{4} = \frac{15 \pm 17}{4} \] Thus: \[ x = \frac{32}{4} = 8 \quad \text{or} \quad x = \frac{-2}{4} = -\frac{1}{2} \] Using \( x = 8 \): 1. First term: \( 8 - 4 = 4 \) 2. Second term: \( 8 + 2 = 10 \) 3. Third term: \( 3(8) + 1 = 25 \) The common ratio \( r \): \[ r = \frac{10}{4} = \frac{5}{2} \] The general term is: \[ a_n = 4 \cdot \left(\frac{5}{2}\right)^{n-1} \] Thus, the general term of the sequence in terms of