A) A.P.
B) G.P.
C) H.P.
D) In G.P. and H.P. both
Correct Answer: C
Solution :
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.
You need to login to perform this action.
You will be redirected in
3 sec
