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

Five numbers are in an $AP$ with a common difference $d \neq 0$. If the $1^{st}$, $3^{rd}$, and $4^{th}$ terms are in a $GP$, then:

Let $a, b$ and $c$ be in $G.P.$ with common ratio $r,$ where $a \ne 0$ and $0 < r \le \frac{1}{2}.$ If $3a, 7b$ and $15c$ are the first three terms of an $A.P.,$ then the $4^{th}$ term of this $A.P.$ is

If $a, b, c$ are in $G.P.$ and $\log a - \log 2b, \log 2b - \log 3c$ and $\log 3c - \log a$ are in $A.P.$,then $a, b, c$ are the lengths of the sides of a triangle which is

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

Let $A_1$ and $A_2$ be two arithmetic means and $G_1, G_2, G_3$ be three geometric means between two distinct positive numbers $a$ and $b$. Then $G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2$ is equal to