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If a, b, c, d and p are different real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0,$ then a, b, c and d are in

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None of these

A.P.

G.P.

H.P.

answer is C.

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Detailed Solution

$(a^{2} p^{2} - 2 a p b + b^{2}) + (b^{2} p^{2} - 2 b p c + c^{2}) + (c^{2} p^{2} - 2 c p d + d^{2}) ⩽ 0 {(a p - b)}^{2} + {(b p - c)}^{2} + {(c p - d)}^{2} \leq 0 \Rightarrow a p = b, b p = c, c p = d \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = p$

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