The product $2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots .$ to $\infty$ is equal to

The sum of an infinite geometric series is $3$. A series, which is formed by squares of its terms, have the sum also $3$. First series will be

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 $c$

A person has $2$ parents, $4$ grandparents, $8$ great grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own.

The sum of the first three terms of a $G.P.$ is $S$ and their product is $27 .$ Then all such $S$ lie in 

Let $a_{1}, a_{2}, a_{3}, \ldots$ be a G.P. such that $a_{1}<0$; $a_{1}+a_{2}=4$ and $a_{3}+a_{4}=16 .$ If $\sum\limits_{i=1}^{9} a_{i}=4 \lambda,$ then $\lambda$ is equal to