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If a,b,c are be in H.P show that a:a-b = a+c : a-c
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@ikram002p
@ganeshie8
@iambatman
If a,b,c are in HP then.... \[\LARGE 2b = \frac{ac}{a+c}\] is it so?
you wrote smt wrong
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Yeah typo wait
a,b,c are in HP so 1/a,1/b,1/c are in AP \[\LARGE \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\] \[\LARGE \frac{2}{b}= \frac{a+c}{ac}\] \[\LARGE b= \frac{2ac}{a+c}\]
yes
To show: \[\LARGE \frac{a}{a-b} = \frac{a+c}{ a-c }\]
\[\LARGE \frac{a}{a-\frac{2ac}{a+c}} = \frac{a+c}{ a-c }\] Take LCM and simplify.. \[\LARGE \frac{a(a+c)}{a^2-{ac}} = \frac{a+c}{ a-c }\] \[\LARGE \frac{\cancel{a}(a+c)}{\cancel{a}(a-{c})} = \frac{a+c}{ a-c }\] hence proved
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oh well, that didn't strike me
just keep following your previous steps
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