SINCE A, B and C are in A.P., therefore \[B=60{}^\circ \]and\[{{b}^{2}}=AC\]. \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2AC}\Rightarrow \frac{1}{2}=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2{{b}^{2}}}\], \[(\because \,{{b}^{2}}=ac)\] \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\Rightarrow {{a}^{2}}+{{c}^{2}}=2{{b}^{2}}\]. "> 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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If the angles \[A,B,C\]of a triangle are in A.P. and the sides \[a,b,c\] opposite to these angles are in G. P. then \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\] are in [MP PET 1998]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 11 months ago

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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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Posted on 22 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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