If $\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^{9}}+\ldots . .+\frac{10240}{3}=2^{ n } \cdot m$, where $m$ is odd, then $m . n$ is equal to

If the sum of an infinite $G.P.$ be $9$ and the sum of first two terms be $5$, then the common ratio is

Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in $G.P.$ Let $b_1=a_1, b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.

$STATEMENT-1$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are neither in $A.P$. nor in $G.P.$ and 

$STATEMENT-2$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are in $H.P.$

If the sum of $n$ terms of a $G.P.$ is $255$ and ${n^{th}}$ terms is $128$ and common ratio is $2$, then first term will be

The sum of infinite terms of a $G.P.$ is $x$ and on squaring the each term of it, the sum will be $y$, then the common ratio of this series is

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