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Question
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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Solution
\[\left( a^2 + b^2 + c^2 \right) p^2 - 2\left( ab + bc + cd \right)p + \left( b^2 + c^2 + d^2 \right) \leq 0\]
\[ \Rightarrow \left( a^2 p^2 + b^2 p^2 + c^2 p^2 \right) - 2\left( abp + bcp + cdp \right) + \left( b^2 + c^2 + d^2 \right) \leq 0\]
\[ \Rightarrow \left( a^2 p^2 - 2abp + b^2 \right) + \left( b^2 p^2 - 2bcp + c^2 \right) + \left( c^2 p^2 - 2cdp + d^2 \right) \leq 0\]
\[ \Rightarrow \left( ap - b \right)^2 + \left( bp - c \right)^2 + \left( cp - d \right)^2 \leq 0\]
\[ \Rightarrow \left( ap - b \right)^2 + \left( bp - c \right)^2 + \left( cp - d \right)^2 = 0\]
\[ \Rightarrow \left( ap - b \right)^2 = 0 \]
\[ \Rightarrow p = \frac{b}{a}\]
\[\text { Also }, \left( bp - c \right)^2 = 0 \]
\[ \Rightarrow p = \frac{c}{b}\]
\[\text { Similiarly }, \Rightarrow \left( cp - d \right)^2 = 0 \]
\[ \Rightarrow p = \frac{d}{c}\]
\[ \therefore \frac{b}{a} = \frac{c}{b} = \frac{d}{c}\]
\[\text { Thus, a, b, c and d are in G . P } .\]
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