A) A. P.
B) G. P.
C) H. P.
D) None of these
Correct Answer: C
Solution :
\[\frac{1}{{{\sin }^{2}}\frac{A}{2}},\frac{1}{{{\sin }^{2}}\frac{B}{2}},\frac{1}{{{\sin }^{2}}\frac{C}{2}}\]are in A. P. Þ \[\frac{1}{{{\sin }^{2}}\frac{C}{2}}-\frac{1}{{{\sin }^{2}}\frac{B}{2}}=\frac{1}{{{\sin }^{2}}\frac{B}{2}}-\frac{1}{{{\sin }^{2}}\frac{A}{2}}\] Þ \[\frac{ab}{(s-a)(s-b)}-\frac{ac}{(s-a)(s-c)}\] \[=\frac{ac}{(s-a)(s-c)}-\frac{bc}{(s-b)(s-c)}\] Þ\[\left( \frac{a}{s-a} \right)\,\left( \frac{b(s-c)-c(s-b)}{(s-b)(s-c)} \right)\]=\[\left( \frac{c}{s-c} \right)\,\left( \frac{a(s-b)-b(s-a)}{(s-a)(s-b)} \right)\] Þ \[abs-abc-acs+abc=acs-abc-bcs+abc\] Þ \[ab-ac=ac-bc\Rightarrow ab+bc=2ac\] or \[\frac{1}{c}+\frac{1}{a}=\frac{2}{b}\], i.e., \[a,b,c\] are in H. P. Note: Students should remember this question as a fact.
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