Question 955913: Find all values of x so that 6x, 4x-1, x^2-1 will be ab arithmetic sequence.
Here is what I did:
6x/a1, 4x-1/a2, x^2-1/a3
a2 - a1 = a3 - a2
(4x-1) - (6x) = (x^2-1) - (4x-1)
4x-1-6x = x^2-1-4x+1
-1-2x = x^2 - 4x
0 = x^2 - 4x +2x + 1
0 = x^2 - 2x + 1
0 = (x-1)(x-1) ..... from here, what do I do now to get the value of x? Am I on the right track?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find all values of x so that 6x, 4x-1, x^2-1 will be ab arithmetic sequence.
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If they form an arithmetic sequence, the pairs must have a common difference.
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Equation:
difference of the 1st pair = difference of the 2nd pair
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(4x-1)-6x = (x^2-1)-(4x-1)
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-2x-1 = x^2-4x
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x^2 - 2x + 1 = 0
Factor:
(x-1)^2 = 0
x = 1
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Cheers,
Stan H.
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