A) \[A.P.\]
B) \[GP\].
C) \[H.P.\]
D) satisfy\[ab=cd\]
Correct Answer: B
Solution :
\[({{a}^{2}}+{{b}^{2}}+{{c}^{2}}){{p}^{2}}-2(ab+bc+cd)p\]\[+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}\le 0\] \[\Rightarrow \]\[({{a}^{2}}{{p}^{2}}-abp+{{b}^{2}})+({{b}^{2}}{{p}^{2}}-2bcp+{{c}^{2}})\] \[\Rightarrow \]\[ap-b=0,\,\,bp-c=0\And cp-d=0\] \[\Rightarrow \]\[\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\Rightarrow a,\,\,b,\,\,c\]and\[d\]in\[GP\] Also\[ad=be\]You need to login to perform this action.
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