Write the first three terms in each of the following sequences defined by the following: $a_{n} = \frac{n-3}{4}$

If the value of $\left(1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots \text{ to } \infty\right)^{\log_{(0.25)}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \text{ to } \infty\right)}$ is $l$,then $l^{2}$ is equal to $......$

Write the first five terms of the sequence defined by $a_{1} = -1$ and $a_{n} = \frac{a_{n-1}}{n}$ for $n \geq 2$,and obtain the corresponding series.

Let $3, 6, 9, 12, \ldots$ up to $78$ terms and $5, 9, 13, 17, \ldots$ up to $59$ terms be two series. Then,the sum of the terms common to both the series is equal to

Let $a_0=0$ and $a_n=3 a_{n-1}+1$ for $n \geq 1$. Then,the remainder obtained by dividing $a_{2010}$ by $11$ is

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: