Define a sequence $\{a_n\}_{n \geq 0}$ by $a_n = \sqrt{\frac{1+a_{n-1}}{2}}$ for $n \geq 1$,with $a_0 = \cos \theta \neq \pm 1$. Then,$\lim_{n \rightarrow \infty} 4^n(1-a_n)$ equals

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 $3, 7, 11, 15, \ldots, 403$ and $2, 5, 8, 11, \ldots, 404$ be two arithmetic progressions. Then the sum of the common terms in them is equal to:

Five numbers are in an $AP$ with a common difference $d \neq 0$. If the $1^{st}$, $3^{rd}$, and $4^{th}$ terms are in a $GP$, then:

Let $a = \min \{x^{2} + 2x + 3 : x \in R\}$ and $b = \lim_{\theta \rightarrow 0} \frac{1 - \cos \theta}{\theta^{2}}$. Then $\sum_{r=0}^{n} a^{r} b^{n-r}$ is

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 $......$