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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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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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Chapter 20: Geometric Progression - Exercise 20.1 [Page 10]

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R.D. Sharma Mathematics [English] Class 11
Chapter 20 Geometric Progression
Exercise 20.1 | Q 15 | Page 10

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