\( 20(k-4) ;(k+1) ; m ; 5 k \) is a set of numbers. The first three terms form an arithmetic sequence and the last three form a geometric sequence. Determine the values of \( k \) and \( m \), if \( k, m>0 \).
Solución de tutoría real
Responder
Solución
¡Inicia sesión para desbloquear respuestas gratis!
Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.
Revisado y aprobado por el equipo de tutoría de UpStudy
Extra Insights
To solve the problem, we start with the conditions given. The first three terms form an arithmetic sequence, so we set up the equation for these terms: \[ (k - 4) + (k + 1) = 2 \cdot 20. \] Simplifying this, we have: \[ 2k - 3 = 40 \implies 2k = 43 \implies k = 21.5. \] Next, we substitute \( k = 21.5 \) into the three terms: \[ 20(k - 4) = 20(21.5 - 4) = 20 \cdot 17.5 = 350, \] \[ k + 1 = 21.5 + 1 = 22.5. \] Now we know the first three terms: \( 350, 22.5, m \). Next, we examine the last three terms, forming a geometric sequence: \[ (k + 1), m, 5k. \] We will use the property of the geometric sequence, where the square of the middle term equals the product of the other two: \[ m^2 = (k + 1)(5k). \] Substituting \( k = 21.5 \): \[ m^2 = (21.5 + 1)(5 \cdot 21.5) = 22.5 \cdot 107.5. \] Calculating \( 22.5 \cdot 107.5 \): \[ 22.5 \cdot 107.5 = 2428.75. \] Thus: \[ m^2 = 2428.75 \implies m = \sqrt{2428.75} \approx 49.28. \] We have determined: \[ k = 21.5, \quad m \approx 49.28. \] Thus, the values of \( k \) and \( m \) are approximately \( k = 21.5 \) and \( m \approx 49.28 \).