A) \[{{b}^{2}}=ac\]
B) \[{{b}^{2}}=\frac{{{a}^{2}}{{c}^{2}}}{{{a}^{2}}+{{c}^{2}}}\]
C) They are in A.P.
D) \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}\]
Correct Answer: C
Solution :
[c] \[\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\Rightarrow \frac{\sin (B+C)}{\sin (A+B)}=\frac{\sin (A-B)}{\sin (B-C)}\] \[\Rightarrow {{\sin }^{2}}B-{{\sin }^{2}}C={{\sin }^{2}}A-{{\sin }^{2}}B\] \[\Rightarrow {{\sin }^{2}}A,{{\sin }^{2}}B,{{\sin }^{2}}C\] and hence \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\] are in A.P.
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