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Question

If a, b, c, d and p are different real numbers such that :

(a2+b2+c2)p22(ab+bc+cd)p+(b2+c2+d2)0, then show that a, b, c and d are in G.P.


Solution

We have,

(a2+b2+c2)p22(ab+bc+cd)p+(b2+c2+d2)0

(a2p22abp+b2)+(b2p22bcp+c2)+(c2p22cdp+d2)0

(apb)2+(bpc)2+(cpd)20

This is only possible when

(apb)2=0p=ba

Also (bpc)2=0p=cb

Similarly, (cpd)2=0p=dc

ba=cb=dc

Hence, a, b, c and d are in G.P.


Mathematics
RD Sharma
Standard XI

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