"For $0 (I) If $\alpha \in(-1,0)$, then $b$ cannot be the geometric mean of $a$ and $c$ (II) If $\alpha \in(0,1)$, then $b$ may be the geometric mean of $a$ and $c$"

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Accepted Answer

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$$\begin{aligned} & f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\ & f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\ & f(1)=0 \\ & \therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c} \\ & \alpha=\frac{c+a-2 b}{a+b-2 c} \\ & \text { If, }-1<\alpha<0 \\ & -1<\frac{c+a-2 b}{a+b-2 c}<0 \\ & b+c<2 a \text { and } b>\frac{a+c}{2} \end{aligned}$$

therefore, b cannot be G.M. between a and c.

$$\begin{aligned} & \text { If, } 0<\alpha<1 \\ & 0<\frac{c+a-2 b}{a+b-2 c}<1 \\ & b>c \text { and } b<\frac{a+c}{2} \end{aligned}$$

Therefore, $\mathrm{b}$ may be the G.M. between $\mathrm{a}$ and $\mathrm{c}$.

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For $0 < c < b < a$, let $(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements

(I) If $\alpha \in(-1,0)$, then $b$ cannot be the geometric mean of $a$ and $c$

(II) If $\alpha \in(0,1)$, then $b$ may be the geometric mean of $a$ and $c$

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For $0 < c < b < a$, let $(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements (I) If $\alpha \in(-1,0)$, then $b$ cannot be the geometric mean of $a$ and $c$ (II) If $\alpha \in(0,1)$, then $b$ may be the geometric mean of $a$ and $c$

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More Sequences and Series questions

Drill 25 more like these. Every day. Free.

For $0 < c < b < a$, let $(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements

(I) If $\alpha \in(-1,0)$, then $b$ cannot be the geometric mean of $a$ and $c$

(II) If $\alpha \in(0,1)$, then $b$ may be the geometric mean of $a$ and $c$