If a,b,c,d and p are distinct real numbers such thatn ( a^2 + b^2 + c^2 )p^2 - 2(ab + bc + cd)p + ( b^2 + c^2 + d^2 ) <= 0n then a,b,c,d

n n (b) : We have (a^2+b^2+c^2) p^2-2(a b+b c+c d) p+(b^2+c^2+d^2) (a^2 p^2-2 a b p+b^2)+(b^2 p^2-2 b c p+c^2)+(c^2 p^2-2 c d p+d^2) (a p-b)^2+(b p-c)^2+(c p-d)^2 alpha=beta=gamma=0)n nnwhich is possible only when a p-b=0, b p-c=0, c p-d=0nnTherefore p=(b / a), p=(c / b), p=(d / c) => (b / a)=(c / b)=(d / c)nn => a, b, c, d are in G.P.

If a,b,c,d and p are distinct real numbers such that ... | Sequence & Series (Mathematics)

\n\begin{aligned}\n& \text { (b) : We have }\left(a^2+b^2+c^2\right) p^2-2(a b+b c+c d) p+\left(b^2+c^2+d^2\right) \leq 0 \\\n& \Rightarrow\left(a^2 p^2-2 a b p+b^2\right)+\left(b^2 p^2-2 b c p+c^2\right)+\left(c^2 p^2-2 c d p+d^2\right) \leq 0 \\\n& \Rightarrow(a p-b)^2+(b p-c)^2+(c p-d)^2 \leq 0 \\\n& (a p-b)^2+(b p-c)^2+(c p-d)^2=0 \\\n& \left(\because \alpha^2+\beta^2+\gamma^2 \leq 0 \Rightarrow \alpha=\beta=\gamma=0\right)\n\end{aligned}\n

\nwhich is possible only when

a p-b=0, b p-c=0, c p-d=0

\n\therefore p=\frac{b}{a}, p=\frac{c}{b}, p=\frac{d}{c} \Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}\n

\Rightarrow a, b, c, d

are in G.P.

Explanation

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