If ${a^2} + a{b^2} + 16{c^2} = 2(3ab + 6bc + 4ac)$, where $a,b,c$ are non-zero numbers. Then $a,b,c$ are in

The third term of a $G.P.$ is the square of first term. If the second term is $8$, then the ${6^{th}}$ term is

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 $c$

If the ratio of the sum of first three terms and the sum of first six terms of a $G.P.$ be $125 : 152$, then the common ratio r is

If $3 + 3\alpha + 3{\alpha ^2} + .........\infty = \frac{{45}}{8}$, then the value of $\alpha $ will be

The $G.M.$ of roots of the equation ${x^2} - 18x + 9 = 0$ is