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 the sum of an infinite $G.P.$,whose first term is $a$ and the common ratio is $r$,be $5$. Let the sum of its first five terms be $\frac{98}{25}$. Then the sum of the first $21$ terms of an $A.P.$,whose first term is $10ar$,$n^{\text{th}}$ term is $a_n$ and the common difference is $10ar^2$,is equal to.

Let $S$ be the sum of the first $9$ terms of the series: $(x+ka) + (x^2+(k+2)a) + (x^3+(k+4)a) + (x^4+(k+6)a) + \ldots$ where $a \neq 0$ and $x \neq 1$. If $S = \frac{x^{10}-x+45a(x-1)}{x-1}$,then $k$ is equal to:

Three numbers form a $G.P.$ If the $3^{rd}$ term is decreased by $64$,the three numbers thus obtained constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$,a $G.P.$ is formed again. Find the numbers.

If $9, x, y, z, a$ are in $A.P.$ such that $x + y + z = 15$,and $9, x, y, z, a$ are in $H.P.$ such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$,then the value of $a$ is:

Three numbers are in an increasing geometric progression with common ratio $r$. If the middle number is doubled,then the new numbers are in an arithmetic progression with common difference $d$. If the fourth term of the $G.P.$ is $3r^{2}$,then $r^{2}-d$ is equal to: