A) \[a,\ b,\ c\]are in A.P.
B) \[{{a}^{2}},\ {{b}^{2}},\ {{c}^{2}}\]are in A.P.
C) \[a,\ b,\ c\]are in G.P.
D) \[a,\ b,\ c\] are in H.P
Correct Answer: D
Solution :
\[2\ln (c-a)=\ln (a+c)+\text{ln}\ (a-2b+c)\] \[\Rightarrow \]\[{{(c-a)}^{2}}=(a+c)(a-2b+c)\] \[\Rightarrow \]\[{{c}^{2}}+{{a}^{2}}-2ac={{(a+c)}^{2}}-2b(a+c)\] \[\Rightarrow \]\[{{c}^{2}}+{{a}^{2}}-2ac={{a}^{2}}+{{c}^{2}}+2ac-2ab-2bc\] \[\Rightarrow \]\[b(a+c)=2ac\]\[\Rightarrow \]\[b=\frac{2ac}{a+c}\]. Hence, \[a,\ b,\ c\]are in H.P.
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