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Mathematics 8 Online
OpenStudy (anonymous):

If a,b,c are be in H.P show that a:a-b = a+c : a-c

OpenStudy (anonymous):

@ikram002p

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

@iambatman

OpenStudy (dls):

If a,b,c are in HP then.... \[\LARGE 2b = \frac{ac}{a+c}\] is it so?

OpenStudy (dls):

you wrote smt wrong

OpenStudy (anonymous):

Yeah typo wait

OpenStudy (dls):

a,b,c are in HP so 1/a,1/b,1/c are in AP \[\LARGE \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\] \[\LARGE \frac{2}{b}= \frac{a+c}{ac}\] \[\LARGE b= \frac{2ac}{a+c}\]

OpenStudy (anonymous):

yes

OpenStudy (dls):

To show: \[\LARGE \frac{a}{a-b} = \frac{a+c}{ a-c }\]

OpenStudy (dls):

\[\LARGE \frac{a}{a-\frac{2ac}{a+c}} = \frac{a+c}{ a-c }\] 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

OpenStudy (anonymous):

oh well, that didn't strike me

OpenStudy (dls):

just keep following your previous steps

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