A) \[~{{b}^{2}}=ac\]
B) \[\frac{c}{a}=\frac{a-b}{b-c}\]
C) \[\frac{a}{c}=\frac{a-b}{b-c}\]
D) \[\frac{a}{b}=\frac{c}{a}\]
Correct Answer: C
Solution :
\[\therefore \,\,\,\,\frac{1}{a},\frac{1}{b},\frac{1}{c}\] are in AP \[\,\,\,\,\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}\Rightarrow \frac{a-b}{ab}=\frac{b-c}{bc}\] \[\Rightarrow \,\,\,\frac{a-b}{a}=\frac{b-c}{c}\Rightarrow \frac{a-b}{b-c}=\]\[\frac{\mathbf{a}}{c}\]
You need to login to perform this action.
You will be redirected in
3 sec
