\( k+1 ; k-1 ; 2 k-5 ; \ldots \) are the first three terms of a geometric sequence, where \( k< \) (a) Calculate the value of \( k \). (b) Determine the sequence. (c) Determine the general term of the sequence. (d) Find the 10 th term.

To solve this problem, we'll start with the properties of a geometric sequence where the ratio between consecutive terms is constant. ### (a) Calculate the value of \( k \). In a geometric sequence, the ratio of the second term to the first term is equal to the ratio of the third term to the second term. Hence, we can write: \[ \frac{k - 1}{k + 1} = \frac{2k - 5}{k - 1} \] Cross multiplying gives: \[ (k - 1)^2 = (k + 1)(2k - 5) \] Expanding both sides: \[ k^2 - 2k + 1 = 2k^2 - 5k + 2k - 5 \] Simplifying yields: \[ k^2 - 2k + 1 = 2k^2 - 3k - 5 \] Rearranging terms leads to: \[ 0 = k^2 - k - 6 \] Factoring gives: \[ 0 = (k - 3)(k + 2) \] Thus, the solutions for \( k \) are \( k = 3 \) or \( k = -2 \). Since \( k < 0 \), we take \( k = -2 \). ### (b) Determine the sequence. Substituting \( k = -2 \) into the terms of the sequence: 1st term: \( k + 1 = -2 + 1 = -1 \) 2nd term: \( k - 1 = -2 - 1 = -3 \) 3rd 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 found by: \[ r = \frac{-3}{-1} = 3 \] The general term \( a_n \) of a geometric sequence can be expressed as: \[ a_n = a_1 \cdot r^{n-1} \] Where \( a_1 = -1 \) and \( r = 3 \). Therefore: \[ a_n = -1 \cdot 3^{n-1} \] ### (d) Find the 10th term. To find the 10th term, we plug \( n = 10 \) into the general term formula: \[ a_{10} = -1 \cdot 3^{10 - 1} = -1 \cdot 3^9 \] Calculating \( 3^9 \): \[ 3^9 = 19683 \] Thus: \[ a_{10} = -19683 \] In summary: - The value of \( k \) is \( -2 \). - The sequence is \( -1, -3, -9 \). - The general term is \( a_n = -1 \cdot 3^{n-1} \). - The 10th term is \( -19683 \).