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x and y are the first and last terms of a sequence of consecutive posi

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Director
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Joined: 07 Dec 2014
Posts: 839

Kudos [?]: 272 [0], given: 15

x and y are the first and last terms of a sequence of consecutive posi [#permalink]

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x and y are the first and last terms of a sequence of consecutive positive odd integers with a range of 22. If y^2-x^2 equals the square of the arithmetic mean of the sequence, what is the sum of x and y?

A. 76
B. 80
C. 84
D. 88
E. 92
[Reveal] Spoiler: OA

Kudos [?]: 272 [0], given: 15

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Manager
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Joined: 24 Jun 2017
Posts: 115

Kudos [?]: 14 [1], given: 126

Re: x and y are the first and last terms of a sequence of consecutive posi [#permalink]

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New post 03 Sep 2017, 16:51
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y - x = 22
y^2-x^2 = (y - x) ( y + x)
(y - x) ( y + x) = mean^2
(y - x) ( y + x) = ((y + x)/2)^2
lets substitute as y-x = 22 and y + x =2x + 22

22(2x+22) = ((2x + 22)/2)^2
simplify
44x + 484 = (x+ 11)^2
44x + 484 = x^2 + 22x + 121

everything to the right
x^2 - 22x - 363 = 0
(x -33) (x+11)
the sequence is a sequence of positive odd integers so x, first term is 33
then y 55 and sum of x + y = 88

Kudos [?]: 14 [1], given: 126

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Manager
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Joined: 27 Dec 2016
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Kudos [?]: 47 [1], given: 205

Concentration: Social Entrepreneurship, Nonprofit
GPA: 3.65
WE: Sales (Consumer Products)
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Re: x and y are the first and last terms of a sequence of consecutive posi [#permalink]

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New post 03 Sep 2017, 17:07
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gracie wrote:
x and y are the first and last terms of a sequence of consecutive positive odd integers with a range of 22. If y^2-x^2 equals the square of the arithmetic mean of the sequence, what is the sum of x and y?

A. 76
B. 80
C. 84
D. 88
E. 92


- We are given set {x,x+2,x+4,....,x+22} which \(x+22=y\) (from the question).
- The number inside the set is 12 (you can write it down one by one from x to x+22).
- So, the mean of the set must be the median, which is the average of 6th and 7th number, or \((x+10)\) and \((x+12)\).

Thus, we can calculate :
- \(y^2-x^2=(mean)^2\)
- Substitute \(y\) with \(x+22\) and mean with the median, we get : \((x+22)^2-x^2=(\frac{(x+10)+(x+12)}{2})^2\)
- Solve the equation, we get two roots of x, \(x=-11\) and \(x=33\). Since the question tells that x is a positive, so we can choose 33.
- If \(33=x\), then \(y=x+22=55\). So \(x+y=88\).

D.
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Kudos [?]: 47 [1], given: 205

Intern
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Joined: 14 Oct 2016
Posts: 30

Kudos [?]: 6 [0], given: 147

Location: India
WE: Sales (Energy and Utilities)
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Re: x and y are the first and last terms of a sequence of consecutive posi [#permalink]

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New post 07 Sep 2017, 11:58
x and y are the first and last terms of a sequence of consecutive positive odd integers with a range of 22. If y^2-x^2 equals the square of the arithmetic mean of the sequence, what is the sum of x and y?

Given the range of the values of x and y is 22.
that means y-x=22
so
y^2 - x^2= {( x+y)/2}^2

because arithmetic mean of even spaced set is also equal to the mean of first and last term of the set.

so
(y-x)(y+x)= {(x+y)(x+y)}/4
so
x+y= 4(y-x)
x+y=88
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Abhimanyu

Kudos [?]: 6 [0], given: 147

Re: x and y are the first and last terms of a sequence of consecutive posi   [#permalink] 07 Sep 2017, 11:58
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