A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
[c] lf \[a,b,c,\]be in H.P. \[\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\Rightarrow b=\frac{2ac}{a+c}\] Now, L.H.S \[=\frac{1}{b-a}+\frac{1}{b-c}\] \[=\frac{1}{\frac{2ac}{a+c}-a}+\frac{1}{\frac{2ac}{a+c}-c}\] \[=\frac{a+c}{2ac-{{a}^{2}}-ca}+\frac{a+c}{2ac-ac-{{c}^{2}}}=\frac{a+c}{ac-{{a}^{2}}}+\frac{a+c}{ac-{{c}^{2}}}\]\[=\frac{a+c}{a(c-a)}+\frac{a+c}{c(c-a)}\] \[=\frac{a+c}{c-a}\left( \frac{1}{a}-\frac{1}{c} \right)=\frac{a+c}{c-a}\left( \frac{c-a}{ac} \right)\] \[=\frac{c+a}{a(c-a)}+\frac{c+a}{c(c-a)}\] Hence, given option [c]be correct.
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