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...
An arithmetic progression is a sequence of numbers where each term is obtained by adding a fixed number to the previous term. For example, the sequence 2, 5, 8, 11, 14 is an arithmetic progression with a common difference of 3.
To find the lowest value for A in an arithmetic progression, you need to know the common difference and at least one term in the sequence. Then, you can use the formula A = a1 - (n-1)d, where A is the lowest value, a1 is the first term, n is the number of terms, and d is the common difference.
Yes, an arithmetic progression can have a negative common difference. This means that the sequence will decrease by a fixed amount each time.
If you only know the first and last term in an arithmetic progression, you can find the common difference by subtracting the first term from the last term. Then, you can use the formula A = a1 - (n-1)d to find the lowest value for A.
No, the formula A = a1 - (n-1)d only applies to arithmetic progressions. For other types of progressions, such as geometric or quadratic progressions, different formulas will need to be used to find the lowest value for A.