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Question

If a,b,c,d and p are distinct real number such that (a2+b2+c2)p22(ab+bc+dc)p+(b2+c2+d2)0, then a,b,c,d :

A
are in A.P.
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B
are in G.P.
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C
are in H.P.
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D
satisfy ab=cd
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Solution

The correct option is B are in G.P.
We have to prove that =ba=cd=dc
Given,(a2+b2+c2)p22(ab+bc+cd)p+(b2+c2+d2)0Proof:a2p2+b2p2+c2p22abp2bcp2cdp+b2+c2+d20(ap)2+b22apbt+(bp)2+(c)22×bp×c+(cp)2+(d)22×cp×d0(apb)2+(bpc)2+(cpd)20(apb)2+(bpc)2+(cpd)20(apb)=0,(bpc)=0&(cpd)=0=L.H.SL.H.S=R.H.SSo,(an+bn).(bn+cn)&(cn+bn)
are in form in G.P

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