Question 983176: Hello,
our topic is all about arithmetic sequence. There's a given question that I found difficulty, so please help me to answer this question:
Find the value of x and y so that (x + y), (3x - 2), (x + 7y), and (3x + 4y) are terms of an arithmetic sequence.
Thank you.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! ok, here we go
arithmetic sequence definition is xn = a + ( d * (n - 1)), where xn is the nth term in the sequence, a is the first term, d is the common difference
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note that x+y is a, then
x1 = x+y
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x2 = 3x-2 = x+y + d
x2 = 2x-y = 2 + d
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x3 = x+7y = x+y + 2d
x3 = 6y = 2d
x3 = y = d/3
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x4 = 3x+4y = x+y + 3d
x4 = 2x+3y = 3d
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using x3 (y = d/3) and x2 we have
2x - d/3 = 2 + d
3x - 2d = 3
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using x3 and x4 we have
2x + d = 3d
x = d, therefore
3x - 2x = 3
x = 3
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now use x4, x=3, d=3
6 + 3y = 9
y = 1
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we have x=3, y=1 and d=3
4, 7, 10, 13
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