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 $c$

Let $a$ and $b$ be roots of ${x^2} - 3x + p = 0$ and let $c$ and $d$ be the roots of ${x^2} - 12x + q = 0$, where $a,\;b,\;c,\;d$ form an increasing G.P. Then the ratio of $(q + p):(q - p)$ is equal to

The number of natural number $n$ in the interval $[1005, 2010]$ for which the polynomial. $1+x+x^2+x^3+\ldots+x^{n-1}$ divides the polynomial $1+x^2+x^4+x^6+\ldots+x^{2010}$ is

If ${G_1}$ and ${G_2}$ are two geometric means and $A$ the arithmetic mean inserted between two numbers, then the value of $\frac{{G_1^2}}{{{G_2}}} + \frac{{G_2^2}}{{{G_1}}}$is

If the sum of first 6 term is $9$ times to the sum of first $3$ terms of the same $G.P.$, then the common ratio of the series will be

Find the sum of the following series up to n terms:

$6+.66+.666+\ldots$