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

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

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03 Sep 2017, 15:49
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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
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Joined: 24 Jun 2017
Posts: 116
Re: x and y are the first and last terms of a sequence of consecutive posi  [#permalink]

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

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07 Sep 2017, 11:58
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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?

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

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14 Sep 2018, 17:45
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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

Since x and y are the first and last terms of a sequence of consecutive positive odd integers with a range of 22, we see that (y - x) = 22. We also know that the arithmetic mean of this evenly-spaced set is (x + y)/2. Since y^2 - x^2 equals the square of the arithmetic mean of the sequence, we have:

y^2 - x^2 = [(x + y)/2]^2

(y + x)(y - x) = (x + y)^2/4

y - x = (x + y)/4

22 = (x + y)/4

88 = x + y

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Re: x and y are the first and last terms of a sequence of consecutive posi   [#permalink] 14 Sep 2018, 17:45
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