For 0(I) If (-1,0), then b cannot be the geometric mean of a and c (II) If (0,1), then b m…

Question

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 $

Accepted Answer

$\begin{gathered} f(x)=(a+b-2c){x}^{2}+(b+c-2a)x+(c+a-2b) \ f(x)=a+b-2c+b+c-2a+c+a-2b=0 \ f(1)=0 \ 	herefore \alpha \cdot 1=\frac{c+a-2b}{a+b-2c} \ \alpha =\frac{c+a-2b}{a+b-2c} \ -1\frac{a+c}{2} \end{gathered}$

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

$ \& 0cbTherefore, $b$ may be the G.M. between $a$ and $

Solutions · 1