Let the first term of a series be $T_1=6$ and its $r^{\text{th}}$ term $T_r=3T_{r-1}+6^r$ for $r=2, 3, \ldots, n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}(n^2-12n+39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$,then $n$ is equal to:

Consider the sequence $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{1}=1, a_{2}=2$ and $a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$ for $n=1, 2, 3, \ldots$. If $\left(\frac{a_{1}+\frac{1}{a_{2}}}{a_{3}}\right) \cdot\left(\frac{a_{2}+\frac{1}{a_{3}}}{a_{4}}\right) \cdot\left(\frac{a_{3}+\frac{1}{a_{4}}}{a_{5}}\right) \cdots\left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=2^{\alpha}\left({}^{61}C_{31}\right)$,then $\alpha$ is equal to.

The product of three consecutive terms of a $G.P.$ is $512$. If $4$ is added to each of the first and the second of these terms,the three terms now form an $A.P.$ Then the sum of the original three terms of the given $G.P.$ is

Let $a_n = \frac{10^n}{n!}$ for $n = 1, 2, 3, \ldots$. Then the greatest value of $n$ for which $a_n$ is the greatest is

If for $x, y \in \mathbb{R}, x > 0,$ $y = \log_{10} x + \log_{10} x^{1/3} + \log_{10} x^{1/9} + \dots$ up to $\infty$ terms and $\frac{2+4+6+\dots+2y}{3+6+9+\dots+3y} = \frac{4}{\log_{10} x}$,then the ordered pair $(x, y)$ is equal to:

The minimum value of the sum of real numbers $a^{-5}, a^{-4}, 3a^{-3}, 1, a^8$ and $a^{10}$ with $a > 0$ is