A) \[a,\ b,\ c\]are in A.P.
B) \[{{a}^{2}},\ {{b}^{2}},\ {{c}^{2}}\]are in A.P.
C) \[{{a}^{2}},\ {{b}^{2}},\ {{c}^{2}}\]are in G. P.
D) None of these
Correct Answer: B
Solution :
\[\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\] \[\Rightarrow \] \[\sin (B+C)\sin (B-C)=\sin (A+B)\sin (A-B)\] \[\Rightarrow \] \[{{\sin }^{2}}B-{{\sin }^{2}}C={{\sin }^{2}}A-{{\sin }^{2}}B\] \[\Rightarrow \] \[2{{\sin }^{2}}B={{\sin }^{2}}A+{{\sin }^{2}}C\Rightarrow 2{{b}^{2}}={{a}^{2}}+{{c}^{2}}\] Hence \[{{a}^{2}},\,{{b}^{2}},\,{{c}^{2}}\] are in A.P.
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