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$5;x;y$ is an arithmetic sequence and $x;y;81$ is a geometric sequence. All terms in the sequences are integers. Calculate the values of $x$ and $y$.
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5;x;y is an arithmetic sequence and x;y;81 is a geometric sequence

x-5 = y-x.......1     and y/x = 81/y.......2 , then solve simultaneously,

from 1, y = 2x-5

subst into 2, (2x-5)/x = 81/(2x-5)

(2x-5)(2x-5) = 81x

4x^2 -10x -10x + 25 = 81x

4x^2 -101x + 25 =0

(4x-1)(x-20) = 0

x = 1/4 or 20

y = 2(20) -5

y = 35

Therefore x = 20 and y = 35 only.
by Diamond (42,406 points)

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