Question 932628: The three terms 7x+1, 9x+7, and 3x-4 (not necessarily in order) are 3 consecutive terms of an arithmetic sequence. Find the largest possible value of the common difference of such an arithmetic sequence.
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website!
Try it all 6 possible ways to order them:
(1) 7x+1, 9x+7, 3x-4
Common difference = (9x+7)-(7x+1) = (3x-4)-(9x+7)
9x+7-7x-1 = 3x-4-9x-7
2x+6 = -6x-11
8x = -17
x = -17/8 = -2.125
Common difference = 2x+6 = 2(-2.125)+6 = 1.75
(2) 7x+1, 3x-4, 9x+7
Common difference = (3x-4)-(7x+1) = (9x+7)-(3x-4)
3x-4-7x-1 = 9x+7-3x+4
-4x-5 = 6x+11
-10x = 16
x = 16/(-10) = -8/5 = -1.6
Common difference = -4x-5 = -4(-1.6)-5 = 1.4
(3) 9x+7, 7x+1, 3x-4
Common difference = (7x+1)-(9x+7) = (3x-4)-(7x+1)
7x+1-9x-7 = 3x-4-7x-1
-2x-6 = -4x-5
2x = 1
x = 1/2 = 0.5
Common difference = -2x-6 = -2(0.5)-6 = -7
(4) 9x+7, 3x-4, 7x+1
Common difference = (3x-4)-(9x+7) = (7x+1)-(3x-4)
3x-4-9x-7 = 7x+1-3x+4
-6x-11 = 4x+5
-10x = 16
x = 16/(-10) = -8/5 = -1.6
x = -8/5 = -1.6
Common difference = -6x-11 = -6(-1.6)-11 = -1.4
(5) 3x-4, 7x+1, 9x+7
Common difference = (7x+1)-(3x-4) = (9x+7)-(7x+1)
7x+1-3x+4 = 9x+7-7x-1
4x+5 = 2x+6
2x = 1
x = 1/2 = 0.5
Common difference = 4x+5 = 4(0.5)+5 = 7
(6) 3x-4, 9x+7, 7x+1
Common difference = (9x+7)-(3x-4) = (7x+1)-(9x+7)
9x+7-3x+4 = 7x+1-9x-7
6x+11 = -2x-6
8x = -17
x = -17/8 = -2.125
Common difference = 6x+11 = 6(-2.125)+11 = -1.75
Case 5 gives the largesr common difference of 7
Edwin
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