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:

Two sequences $\{t_n\}$ and $\{s_n\}$ are defined by $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right)$ and $s_n = \left[ \log \left( \frac{5}{3} \right) \right]^n$. Then:

Let $a_n = \frac{10^n}{n!}$ for $n = 1, 2, 3, \ldots$. The greatest value of $n$ for which $a_n$ is the greatest is:

The value of $\left[\frac{2^{2020}+1}{2^{2018}+1}\right]+\left[\frac{3^{2020}+1}{3^{2018}+1}\right]+\left[\frac{4^{2020}+1}{4^{2018}+1}\right] +\left[\frac{5^{2020}+1}{5^{2018}+1}\right] + \left[\frac{6^{2020}+1}{6^{2018}+1}\right]$ is (where $[\cdot]$ denotes the greatest integer function):

Suppose the sequence $a_1, a_2, a_3, \ldots$ is an arithmetic progression of distinct numbers such that the sequence $a_1, a_2, a_4, a_8, \ldots$ is a geometric progression. The common ratio of the geometric progression is

If the sum of the first $40$ terms of the series $3+4+8+9+13+14+18+19+\ldots$ is $(102)m$,then $m$ is equal to: