A) are in AP
B) are in GP
C) are in HP
D) satisfy a + b = c
Correct Answer: A
Solution :
Using formula \[\cos \left( \frac{C}{2} \right)=\sqrt{\frac{s(s-c)}{ab}}\] and \[\cos \left( \frac{A}{2} \right)=\sqrt{\frac{s(s-a)}{bc}}\]. Given, \[a{{\cos }^{2}}\left( \frac{C}{2} \right)+c{{\cos }^{2}}\left( \frac{A}{2} \right)=\frac{3\,b}{2}\] \[\Rightarrow a\left[ \frac{s\,(s-c)}{ab} \right]+c\left[ \frac{s\,(s-a)}{bc} \right]=\frac{3\,b}{2}\] \[\Rightarrow \] \[\frac{s\,(s-c+s-a)}{b}=\frac{3b}{2}\] \[\Rightarrow \] \[2s\,(2s-c-a)=3{{b}^{2}}\] \[\Rightarrow \] \[2s(a+b+c-c-a)=3{{b}^{2}}\] \[\Rightarrow \] \[(a+b+c)b=3{{b}^{2}}\] \[\Rightarrow \] \[a+b+c=3b\] \[\Rightarrow \] \[2b=a+c\] Hence, a, b, c are in AP.
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