A) G.P.
B) A. P.
C) H.P.
D) None of these
Correct Answer: B
Solution :
[b] \[\therefore cotA,cotB,cotC\text{ }be\text{ }in\text{ }A.\text{ }P.\] Then \[2.cotB=cotA+cotC\] \[\Rightarrow 2\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}+\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}\]\[\Rightarrow 2\frac{s(s-b)}{\Delta }=\frac{s(s-a)}{\Delta }+\frac{s(s-c)}{\Delta }\] \[\Rightarrow 2s\left( s-b \right)=s\left( s-a \right)+s\left( s-c \right)~\] \[\Rightarrow 2\left( s-b \right)=\left( s-a \right)+\left( s-c \right)~\] \[\Rightarrow 2s-2b=2s-\left( a+c \right)\] \[\Rightarrow 2b=a+c\] Hence \[a,b,c\]be in A.P.
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