A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
[c] : Let \[a=\tan x-\tan y,b=\tan y-\tan z\]and \[c=\tan z-\tan x\] \[\therefore \]a + b + c = 0 ...(i) From (i), \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}+2ac\] ...(ii) According to question, \[2{{b}^{2}}={{a}^{2}}+{{c}^{2}}\] \[\Rightarrow \]\[2{{b}^{2}}={{b}^{2}}-2ac\] [Using (ii)] \[\Rightarrow \]\[-{{b}^{2}}=2ac\] \[\Rightarrow \]\[-b=\frac{2ac}{b}=\frac{2ac}{-(a+c)}\Rightarrow b=\frac{2ac}{a+c}\] \[\therefore \]a, b, c are in H.P.
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