A) a,b,c are in AP
B) \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\]are in AP
C) a,b, care in GP
D) \[a,b,c\]careinHP
Correct Answer: D
Solution :
Given that\[\log (a+c),\log (c-a),\log (a-2b+c)\]are in AP. \[\therefore \]\[2\text{ }log(c-a)=log(a+c)+log(a-2b+c)\] \[\Rightarrow \] \[log{{(c-a)}^{2}}=log\{(a+c)(a-2b+c)\}\] \[\Rightarrow \]\[{{c}^{2}}+{{a}^{2}}-2ac={{a}^{2}}-2ab+ac+ac-2bc+{{c}^{2}}\] \[\Rightarrow \] \[-4ac=-2(ab+bc)\] On dividing by \[-2abc,\] \[\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\] Thus, a,b,c are in HP.
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