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

Let $729, 81, 9, 1, \dots$ be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence. If $2\sum_{n=1}^{40}(P_{n})^{\frac{1}{n}}=\frac{3^{\alpha}-1}{3^{\beta}}$ and $\gcd(\alpha,\beta)=1$, then $\alpha+\beta$ is equal to

In a $G.P.$,the $3^{rd}$ term is $24$ and the $6^{th}$ term is $192$. Find the $10^{th}$ term.

The arithmetic mean of two numbers $b$ and $c$ is $a$,and $g_1$ and $g_2$ are two geometric means between them. If $g_1^3 + g_2^3 = kabc$,then $k = \dots$

Consider two $G$.Ps. $2, 2^{2}, 2^{3}, \ldots$ and $4, 4^{2}, 4^{3}, \ldots$ of $60$ and $n$ terms respectively. If the geometric mean of all the $60+n$ terms is $(2)^{\frac{225}{8}}$,then $\sum_{k=1}^{n} k(n-k)$ is equal to.

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