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Senior Manager
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In a set of integers, each term is two more than the [#permalink]
 19 Feb 2004, 08:47
In a set of integers, each term is two more than the previous term and x, y, and z are the first, middle and last terms, respectively. which of the following equals x^2 + 2xz + z^2
a. 2y
b. 2y^2
c. 4y
d. 4y^2
e. 6y^4
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Intern
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I am new member here, this is my first attempt. Hope to contribute and learn.
x^2+2xz+z^2 = (x+z)^2 = (x+z)(x+z)
Y is the middle term, so
(x+z) = y-2 + y+2 = 2y
Thus answer is
2y.2y = 4y^2
Thanks.
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Senior Manager
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Are they Consecutive integers??? 
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Intern
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Since they did not mention how many are there in the series, I assumed smallest possible series namely, 3 (x, y, z). Could be wrong.
-SP
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SVP
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Exactly same procedure as of asagem99
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Director
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Agree with the crowd!
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