In an increasing geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is $25.$ Then, the sum of $4^{th},6^{th}$ and $8^{th}$ terms is equal to:

The sum of the first three terms of $\mathrm{G}.\mathrm{P}$ is $S$and their products is $27$. Then all such $S$ lie in

Let $S$ be the sum of the first $9$ term of the series :

$\left\{x+ka\right\}+\left\{x^2+\left(k+2\right)a\right\}+\left\{x^3+\left(k+4\right)a\right\}+\left\{x^4+\left(k+6\right)a\right\}+\dots$ where $a\ne 0$ and $x \ne 1$. If $S=\frac{x^{10}-x+45a\left(x-1\right)}{x-1}$ , then $k$ is equal to