A) a, b, c, d are in A.P.
B) ab = cd
C) ac = bd
D) a, b, c, d are in G.P.
Correct Answer: D
Solution :
The given relation can be written as \[({{a}^{2}}{{p}^{2}}-2abp+{{b}^{2}})+({{b}^{2}}{{p}^{2}}+{{c}^{2}}-2bpc)+\] \[({{a}^{2}}{{p}^{2}}+{{d}^{2}}-2pcd)\le 0\] or\[{{(ap-b)}^{2}}+{{(bp-c)}^{2}}+{{(cp-d)}^{2}}\le 0\] ...(1) Since a, b, c, d and p are all real, the inequality is possible only when each of factor is zero. i.e., ap - b = 0, bp - c = 0 and cp - d = 0 or \[p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\]or a,b,c,d are in G.P.
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