Let $\alpha = \sum_{n=101}^{200} 2^n \sum_{k=101}^n \frac{1}{k !}$ and $b = \sum_{n=101}^{200} \frac{2^{201}-2^n}{n !}$. Then,$\frac{a}{b}$ is

Let $S_{n}(x) = \log_{a^{1/2}} x + \log_{a^{1/3}} x + \log_{a^{1/6}} x + \log_{a^{1/11}} x + \log_{a^{1/18}} x + \log_{a^{1/27}} x + \ldots$ up to $n$-terms,where $a > 1$. If $S_{24}(x) = 1093$ and $S_{12}(2x) = 265$,then the value of $a$ is equal to ..... .

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:

Find the $7^{\text{th}}$ term in the following sequence whose $n^{\text{th}}$ term is $a_{n} = \frac{n^{2}}{2^{n}}$.

If all the terms of an $A.P.$ are squared,then the new series will be in

Let four distinct positive numbers $a_1, a_2, a_3, a_4$ be in a geometric progression. 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-$I$: The numbers $b_1, b_2, b_3, b_4$ are neither in an arithmetic progression nor in a geometric progression.
Statement-$II$: The numbers $b_1, b_2, b_3, b_4$ are in a harmonic progression.