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

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VP
Joined: 07 Dec 2014
Posts: 1111
The first term of a sequence of consecutive positive even integers is  [#permalink]

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03 Jun 2017, 09:09
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Difficulty:

35% (medium)

Question Stats:

73% (02:40) correct 27% (03:35) wrong based on 58 sessions

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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
Intern
Joined: 25 Dec 2016
Posts: 18
Re: The first term of a sequence of consecutive positive even integers is  [#permalink]

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03 Jun 2017, 09:23
1
Calculate through POE
And first term is 8
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Joined: 26 Feb 2016
Posts: 3302
Location: India
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The first term of a sequence of consecutive positive even integers is  [#permalink]

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03 Jun 2017, 09:27
2
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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Location: India
Re: The first term of a sequence of consecutive positive even integers is  [#permalink]

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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
VP
Joined: 07 Dec 2014
Posts: 1111
The first term of a sequence of consecutive positive even integers is  [#permalink]

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03 Jun 2017, 10:13
2
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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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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