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

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

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

The $5^{\text {th }}, 8^{\text {th }}$ and $11^{\text {th }}$ terms of a $G.P.$ are $p, q$ and $s,$ respectively. Show that $q^{2}=p s$

If $a, b, c$ and $d$ are in $G.P.$ show that:

$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$

If the $n^{th}$ term of geometric progression $5, - \frac{5}{2},\frac{5}{4}, - \frac{5}{8},...$ is $\frac{5}{{1024}}$, then the value of $n$ is

If $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}(x \neq 0),$ then show that $a, b, c$ and $d$ are in $G.P.$

If ${\log _a}x,\;{\log _b}x,\;{\log _c}x$ be in $H.P.$, then $a,\;b,\;c$ are in