A) A. P.
B) H. P.
C) G. P.
D) None of these
Correct Answer: A
Solution :
Since A, B and C are in A.P., therefore \[B=60{}^\circ \]and\[{{b}^{2}}=ac\]. \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\Rightarrow \frac{1}{2}=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2{{b}^{2}}}\], \[(\because \,{{b}^{2}}=ac)\] \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\Rightarrow {{a}^{2}}+{{c}^{2}}=2{{b}^{2}}\].
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