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Q.

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

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a

are in H.P.

b

are in A. P

c

are in G.P. 

d

satisfy  ab=cd

answer is B.

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Detailed Solution

(a2p22abp+b2)+(b2p22bcp+c2)  +  (c2p22cdp+d2)0
(apb)2+(bpc)2+(cpd)20           
(apb)2+(bpc)2+(cpd)2=0            is possible only 

 If  apb=0, bpc=0,cpd=0

  ba=p,cb=p,dc=p  ba=cb=dc=p
           
            

            
 

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If  a,b,c,d  and  p are distinct real numbers such that   (a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)≤0   then  a,b,c,d .