A) A.P
B) G.P
C) H.P
D) None of these
Correct Answer: B
Solution :
\[(a+2b+2c)\]\[(a-2b+2c)={{a}^{2}}+4{{c}^{2}}\] Þ \[{{(a+2c)}^{2}}-{{(2b)}^{2}}={{a}^{2}}+4{{c}^{2}}\] Þ \[{{a}^{2}}+4ac+4{{c}^{2}}-4{{b}^{2}}={{a}^{2}}+4{{c}^{2}}\] Þ \[4ac-4{{b}^{2}}=0\]Þ \[{{b}^{2}}=ac\] Hence a, b, c are in G.P.
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