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Standard X

Mathematics

General Form of an AP

If a,b,c, are...

Question

If a,b,c, are in G.P. and x,y respectively arithmetic means between a,b and b,c then prove that
$\frac{1}{x} + \frac{1}{y} = \frac{2}{b}$

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Solution

$\frac{1}{x} + \frac{1}{y} = \frac{2}{a + b} + \frac{2}{b + c} = \frac{2 (a + c + 2 b)}{a b + a c + b^{2} + b c}$
put $b^{2} = a c$
$= \frac{2 (a + c + 2 b)}{a b + 2 b^{2} + b c} = \frac{2 (a + c + 2 b)}{b (a + c + 2 b)} = \frac{2}{b}$

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If a,b,c, are in G.P. and x,y respectively arithmetic means between a,b and b,c then prove that1x+1y=2b

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If a,b,c, are in G.P. and x,y respectively arithmetic means between a,b and b,c then prove that
$\frac{1}{x} + \frac{1}{y} = \frac{2}{b}$

$\frac{1}{x} + \frac{1}{y} = \frac{2}{a + b} + \frac{2}{b + c} = \frac{2 (a + c + 2 b)}{a b + a c + b^{2} + b c}$
put $b^{2} = a c$
$= \frac{2 (a + c + 2 b)}{a b + 2 b^{2} + b c} = \frac{2 (a + c + 2 b)}{b (a + c + 2 b)} = \frac{2}{b}$