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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.
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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TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Assume 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.

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

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I thought about the inequality matter. But we will get periods.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1solve the 3 inequalities and take the intersection

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I think according to your assumption , we can take (a) as common factor, can't we ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1divide both sides by \(a\) so that it disappears

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1you can do so because \(a\) is assumed to be side of a triangle, so \(a\gt 0\)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0If inequalities really help, why don't we use GM AM inequality , xy = z^2 and the triangle inequality. \[(r\frac{ 1  \sqrt(5) }{ 2 })( r  \frac{ 1 + \sqrt(5) }{ 2 }) > \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1use it if you find it simpler than triangle inequality

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Not easier at all. Imagine z + y + x > 3root(zyz)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1triangle inequlity does the job here

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=solve+r%5E2%3C1%2Br%2C+1%3Cr%2Br%5E2%2C+r%3C1%2Br%5E2

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0what 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

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0assume x, y , z = a, ar, ar^2 and use this with your inequality and maybe we will reach something.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{\phi} \lt r\lt \phi\] where \(\phi\) = \(\href{https:///en.wikipedia.org/wiki/Golden_ratio}{\text{golden ratio}}\)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I read about the golden ration but it's over my post level.
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