A) In H.P.
B) In G.P.
C) In A.P.
D) None of these
Correct Answer: A
Solution :
We have \[\frac{1}{a}+\frac{1}{c}+\frac{1}{a-b}+\frac{1}{c-b}=0\] \[\frac{1}{a}+\frac{1}{c-b}=\frac{1}{b-a}-\frac{1}{c}\] \[\Rightarrow \] \[\frac{c-b+a}{a(c-b)}=\frac{c-b+a}{(b-a)c}\]\[\Rightarrow \]\[ac-ab=bc-ac\] \[\Rightarrow \]\[2ac=ab+bc\]\[\Rightarrow \]\[\frac{2ac}{a+c}=b\]i.e., \[a,b,c\] are in H.P.You need to login to perform this action.
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