@Kajal...nishant did it correctly....
If a,b ,c,d and p are different real numbers such that :(a2+b2+c2)p2-2(ab+bc+cd)p+(b2+c2+d2)≤0, then show that a,b,c and d are in G.P
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2 Answers
Lokesh Verma
·2010-10-04 02:25:54
(ap-b)2+(bp-c)2+(cp-d)2≤0
But sum of three squares..
so ap=b, bp=c and cp=d
so they are in GP