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:

For every natural number $n$,$n(n + 1)$ is always

If $m$ is the $A.M.$ of two distinct real numbers $l$ and $n$ $(l, n > 1)$ and $G_1, G_2, G_3$ are three geometric means between $l$ and $n$,then $G_1^4 + 2G_2^4 + G_3^4$ equals:

If $|x| < 1, |y| < 1$ and $x \neq y,$ then the sum to infinity of the following series $(x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$ is:

If $[x]$ represents the greatest integer not greater than $x$,then $\left[\left(1+\frac{1}{100000}\right)^{100000}\right]=$

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: