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The first term of a sequence of consecutive positive even integers is

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The first term of a sequence of consecutive positive even integers is  [#permalink]

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New post 03 Jun 2017, 09:09
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The first term of a sequence of consecutive positive even integers is equal to half the range. If the difference between the squares of the first and last terms is 512, what is the sum of the first and last terms?

A. 24
B. 28
C. 32
D. 40
E. 44
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Re: The first term of a sequence of consecutive positive even integers is  [#permalink]

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New post 03 Jun 2017, 09:23
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Answer C
Calculate through POE
And first term is 8
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The first term of a sequence of consecutive positive even integers is  [#permalink]

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New post 03 Jun 2017, 09:27
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Let the first term of the sequence be a, and the last term is b

Given: \(a = \frac{(b-a)}{2}\) -> \(b - a = 2a\)

\(b^2 - a^2 = 512\)
\((b - a)(b + a) = 512\)
\(2a*(b+a) = 512\) (because b - a = 2a)
\(a * (b+a) = 256\)

The only number which divides 256\((2^8)\) perfectly is 32\((2^5)\) which is the sum of first and last terms (Option C)
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Re: The first term of a sequence of consecutive positive even integers is  [#permalink]

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New post 03 Jun 2017, 09:54
Let first term be y and last term be x. Given: x^2 - y^2 = 512 or (x+y)(x-y) = 512
Now range = (x-y) = twice of first term (given) = 2y

So, (x+y)*2y = 512 or (x+y) = 256/y

Since y is a positive even integer, 256 must be divisible by y and so 256/y must give an integer as a result - and that result must be a factor of 256. So (x+y) must be a factor of 256.

Out of given options, only 32 is a factor of 256. Hence C answer
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The first term of a sequence of consecutive positive even integers is  [#permalink]

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New post 03 Jun 2017, 10:13
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gracie wrote:
The first term of a sequence of consecutive positive even integers is equal to half the range. If the difference between the squares of the first and last terms is 512, what is the sum of the first and last terms?

A. 24
B. 28
C. 32
D. 40
E. 44


another approach:
let x and y=first and last terms
y=x+2x=3x
9x^2-x^2=512
8x^2=512
x^2=64
x=8
8+2*8=24
8+24=32
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Re: The first term of a sequence of consecutive positive even integers is  [#permalink]

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New post 14 Sep 2018, 03:10
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Re: The first term of a sequence of consecutive positive even integers is &nbs [#permalink] 14 Sep 2018, 03:10
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