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

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Senior Manager
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In a set of integers, each term is two more than the [#permalink] New post 19 Feb 2004, 07:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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 [#permalink] New post 19 Feb 2004, 08:07
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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 [#permalink] New post 19 Feb 2004, 08:19
Are they Consecutive integers??? :?
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 [#permalink] New post 19 Feb 2004, 08:29
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.

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 [#permalink] New post 19 Feb 2004, 08:42
Exactly same procedure as of asagem99
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 [#permalink] New post 19 Feb 2004, 11:37
Agree with the crowd!
  [#permalink] 19 Feb 2004, 11:37
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In a set of integers, each term is two more than the

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