A) H. P.
B) G. P.
C) A. P.
D) None of these
Correct Answer: C
Solution :
\[\cot A,\cot B\]and \[\cot C\]are in A. P. Þ \[\cot A+\cot C=2\cot B\] Þ \[\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}=\frac{2\cos B}{\sin B}\] Þ \[\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc(ka)}+\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab(kc)}=2\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac(kb)}\] Þ \[{{a}^{2}}+{{c}^{2}}=2{{b}^{2}}\]. Hence\[{{a}^{2}},{{b}^{2}},{{c}^{2}}\]are in A. P. Note : Students should remember this question as a fact.
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