The discussion focuses on finding the lowest value of the expression A = a1a2 + a2a3 + a3a1 for an arithmetic progression defined by a1, a2, a3, and a fourth term of 4 with a common difference d. The initial approach led to the equation A = 3x² + 6xd + 2d², where a1 = x and a2 = x + d. A key insight was provided, emphasizing the importance of incorporating the fourth term to simplify A into a single-variable quadratic equation, which allows for easier minimization.
PREREQUISITESMathematicians, students studying algebra, and anyone interested in optimization techniques related to arithmetic progressions.
Hi mitaka90!mitaka90 said:a1, a2, a3 and 4 make an arithmetic progression with difference d. For which values of d, A = a1a2 + a2a3 + a3a1 has the lowest value?I don't know if I went with the right approach, but I managed to get this : A=3x2 +6xd + 2d2 for a1= x, a2 = x + d, etc... But I don't know what else to do.
anemone said:Hi mitaka90!
It seems to me you haven't made good use of the given fourth term in that arithmetic progression...:) the fourth term would help you to simplify your $A$ in terms of only one variable and when you have the quadratic equation in terms of one variable, I believe you could handle from there...