MHB Find Lowest Value for A: a1, a2, a3 & 4 | Arithmetic Progression

mitaka90
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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.
 
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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.
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...
 
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...

Omg, I'm such a moron. I hate it when I do the hard work and then the easiest and most noticable thing just slips from my sight. Thank you sincerely, I guess that little tip is what I needed.
 

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