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# In a set of integers, each term is two more than the

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Senior Manager
Joined: 19 Feb 2004
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Location: Lungi
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In a set of integers, each term is two more than the [#permalink]  19 Feb 2004, 07:47
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
Intern
Joined: 26 Jan 2004
Posts: 39
Location: US
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Kudos [?]: 0 [0], given: 0

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
2y.2y = 4y^2

Thanks.
Senior Manager
Joined: 05 Feb 2004
Posts: 290
Location: USA
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Kudos [?]: 3 [0], given: 0

Are they Consecutive integers???
Intern
Joined: 26 Jan 2004
Posts: 39
Location: US
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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
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
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Exactly same procedure as of asagem99
Director
Joined: 03 Jul 2003
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Agree with the crowd!
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