RECIPROCALS of \[a\] and \[c\] occur on RHS, let us first assume that \[a,\ b,\ c\] are in H.P. So that \[\FRAC{1}{a},\ \frac{1}{b},\ \frac{1}{c}\] are in A.P. \[\Rightarrow \] \[\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}=d\], say \[\Rightarrow \]\[\frac{a-b}{ab}=d=\frac{b-c}{bc}\Rightarrow a-b=ABD\] and \[b-c=bcd\]  Now LHS \[=-\frac{1}{a-b}+\frac{1}{b-c}=-\frac{1}{abd}+\frac{1}{bcd}\] \[=\frac{1}{bd}\left( \frac{1}{c}-\frac{1}{a} \RIGHT)=\frac{1}{bd}(2D)\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}=\]RHS \[\therefore \ \ a,.\ b,\ c\] are in H.P. is verified.

"> RECIPROCALS of \[a\] and \[c\] occur on RHS, let us first assume that \[a,\ b,\ c\] are in H.P. So that \[\FRAC{1}{a},\ \frac{1}{b},\ \frac{1}{c}\] are in A.P. \[\Rightarrow \] \[\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}=d\], say \[\Rightarrow \]\[\frac{a-b}{ab}=d=\frac{b-c}{bc}\Rightarrow a-b=ABD\] and \[b-c=bcd\]  Now LHS \[=-\frac{1}{a-b}+\frac{1}{b-c}=-\frac{1}{abd}+\frac{1}{bcd}\] \[=\frac{1}{bd}\left( \frac{1}{c}-\frac{1}{a} \RIGHT)=\frac{1}{bd}(2D)\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}=\]RHS \[\therefore \ \ a,.\ b,\ c\] are in H.P. is verified.

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If \[\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}\], then \[a,\ b,\ c\]  are in [MNR 1984; MP PET 1997; UPSEAT 2000]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 9 months ago

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Since the RECIPROCALS of \[a\] and \[c\] occur on RHS, let us first assume that \[a,\ b,\ c\] are in H.P. So that \[\FRAC{1}{a},\ \frac{1}{b},\ \frac{1}{c}\] are in A.P. \[\Rightarrow \] \[\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}=d\], say \[\Rightarrow \]\[\frac{a-b}{ab}=d=\frac{b-c}{bc}\Rightarrow a-b=ABD\] and \[b-c=bcd\]  Now LHS \[=-\frac{1}{a-b}+\frac{1}{b-c}=-\frac{1}{abd}+\frac{1}{bcd}\] \[=\frac{1}{bd}\left( \frac{1}{c}-\frac{1}{a} \RIGHT)=\frac{1}{bd}(2D)\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}=\]RHS \[\therefore \ \ a,.\ b,\ c\] are in H.P. is verified.

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