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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
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The correct answer is:
H.P.
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.
$\Rightarrow \quad \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}=d$, say
$\Rightarrow \frac{a-b}{a b}=d=\frac{b-c}{b c} \Rightarrow a-b=a b d$ and $b-c=b c d$
Now LHS $=-\frac{1}{a-b}+\frac{1}{b-c}=-\frac{1}{a b d}+\frac{1}{b c d}$
$=\frac{1}{b d}\left(\frac{1}{c}-\frac{1}{a}\right)=\frac{1}{b d}(2 d) \Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}={ }_{\text {RHS }}$
$\therefore a, b, c$ are in H.P. is verified.
So that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.
$\Rightarrow \quad \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}=d$, say
$\Rightarrow \frac{a-b}{a b}=d=\frac{b-c}{b c} \Rightarrow a-b=a b d$ and $b-c=b c d$
Now LHS $=-\frac{1}{a-b}+\frac{1}{b-c}=-\frac{1}{a b d}+\frac{1}{b c d}$
$=\frac{1}{b d}\left(\frac{1}{c}-\frac{1}{a}\right)=\frac{1}{b d}(2 d) \Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}={ }_{\text {RHS }}$
$\therefore a, b, c$ are in H.P. is verified.
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