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24 Jan 2010 08:43:49 IST

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If a,b,c are in H.P. then prove that a/(b+c), b/(c+a), c/(a+b) are in Harmonic Progression.

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If a,b,c are in H.P. then prove that a/(b+c), b/(c+a), c/(a+b) are in Harmonic Progression.

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Abhishek S

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24 Jan 2010 09:09:46 IST

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For that u should prove that b/c+a is the harmonic mean of a/b+c and c/a+b.

now 2 (ac/(a+b)(b+c))/a^a+ab+bc+c^2

solve the above and prove that it quals b/c+a

jee-do or die

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24 Jan 2010 09:30:53 IST

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since a

a, b , c are in H.P

we know that b = 2ac / a+ c

now we have to prove that a/ b+ c , b / c+ a , c / a + b are in h.p

i .e we have to prove that b = 2ac / a+ c ---1

$\frac{b}{c + a} = 2\left ( \frac{\frac{a}{b+ c .}\frac{c}{a+ b} }{\frac{a}{b+ c}+ \frac{c}{a+ b}} \right )$

$\frac{b}{c + a} = \frac{2ac}{a^2 + ab + bc + c^2 }$

$\frac{b}{c + a} = \frac{2ac}{a^2 + c^2 + b ( a+ c ) }$

substitute value of a+ c from --1 in above equation

$\frac{b}{c + a} = \frac{2ac}{a^2 + c^2 + b (\frac{2ac}{b}) }$

$\frac{b}{c + a} = \frac{2ac}{a^2 + c^2 + 2ac }$

$\frac{b}{c + a} = \frac{2ac}{(a+ c)^{2} }$

$b = \frac{2ac}{a+ c}$

hence proved these are in h.p

Hari Shankar

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24 Jan 2010 09:59:35 IST

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Just remember this: if x,y,z are in AP, so are x+k,y+k,z+k and kx,ky, kz for any number k

a,b,c are in HP

$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP,

$\Rightarrow \frac{a+b+c}{a}-1, \frac{a+b+c}{b}-1, \frac{a+b+c}{c}-1$ are in AP

$\Rightarrow \frac{b+c}{a}, \frac{a+c}{b}, \frac{a+b}{c}$ are in AP

$\Rightarrow \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in HP

pvsaratchandra

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25 Jan 2010 20:15:09 IST

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Thanks for your solution & immediate reply. Thanks again.

- Sarat Chandra, Bangalore

abdulfukar

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17 May 2012 15:12:21 IST

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If a,b,c are in H.P. then prove that a/(b+c), b/(c+a), c/(a+b) are in Harmonic Progression.

abdulfukar

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17 May 2012 15:12:37 IST

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If a,b,c are in H.P. then prove that a/(b+c), b/(c+a), c/(a+b) are in Harmonic Progression.

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