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

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