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