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TrojanPoem

  • one year ago

If three consecutive terms for a geometric sequence formed a triangle sides.what is the base for the geometric sequence.

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  1. TrojanPoem
    • one year ago
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    Assume the terms are x, y, z. As they form a geometric sequence ( x, y, z) zx = y^2 -> (1) Now we need 2 more equations.

  2. ganeshie8
    • one year ago
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    Let the terms be \(a, ~ar,~ ar^2 \) For these to be sides of a triangle, they must satisfy the triangle inequality relation : \[a+ar \gt ar^2\tag{1}\] \[ar+ar^2 \gt a\tag{2}\] \[ar^2+a \gt ar\tag{3}\]

  3. TrojanPoem
    • one year ago
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    I thought about the inequality matter. But we will get periods.

  4. ganeshie8
    • one year ago
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    solve the 3 inequalities and take the intersection

  5. TrojanPoem
    • one year ago
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    I think according to your assumption , we can take (a) as common factor, can't we ?

  6. ganeshie8
    • one year ago
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    divide both sides by \(a\) so that it disappears

  7. ganeshie8
    • one year ago
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    you can do so because \(a\) is assumed to be side of a triangle, so \(a\gt 0\)

  8. TrojanPoem
    • one year ago
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    If inequalities really help, why don't we use GM AM inequality , xy = z^2 and the triangle inequality. \[(r-\frac{ 1 - \sqrt(5) }{ 2 })( r - \frac{ 1 + \sqrt(5) }{ 2 }) > \]

  9. ganeshie8
    • one year ago
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    use it if you find it simpler than triangle inequality

  10. TrojanPoem
    • one year ago
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    Not easier at all. Imagine z + y + x > 3root(zyz)

  11. ganeshie8
    • one year ago
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    triangle inequlity does the job here

  12. ganeshie8
    • one year ago
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    http://www.wolframalpha.com/input/?i=solve+r%5E2%3C1%2Br%2C+1%3Cr%2Br%5E2%2C+r%3C1%2Br%5E2

  13. TrojanPoem
    • one year ago
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    what about this ? zy = x^2 z + y + x > 3 3root(xyz) from (1) z + y+ x > 3 3root(x^3) z +y + x > 3 x z + y > 2x

  14. TrojanPoem
    • one year ago
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    assume x, y , z = a, ar, ar^2 and use this with your inequality and maybe we will reach something.

  15. ganeshie8
    • one year ago
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    we're done.

  16. TrojanPoem
    • one year ago
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    Really ? Xd

  17. TrojanPoem
    • one year ago
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    brb

  18. ganeshie8
    • one year ago
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    \[\frac{1}{\phi} \lt r\lt \phi\] where \(\phi\) = \(\href{https:///en.wikipedia.org/wiki/Golden_ratio}{\text{golden ratio}}\)

  19. TrojanPoem
    • one year ago
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    Ok, I read about the golden ration but it's over my post level.

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