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 $V_r$ denote the sum of the first $r$ terms of an arithmetic progression $(A.P.)$ whose first term is $r$ and the common difference is $(2r-1)$. Let $T_r = V_{r+1} - V_r - 2$ and $Q_r = T_{r+1} - T_r$ for $r = 1, 2, \ldots$
$1.$ The sum $V_1 + V_2 + \ldots + V_n$ is
$(A)$ $\frac{1}{12} n(n+1)(3n^2-n+1)$
$(B)$ $\frac{1}{12} n(n+1)(3n^2+n+2)$
$(C)$ $\frac{1}{2} n(2n^2-n+1)$
$(D)$ $\frac{1}{3}(2n^3-2n+3)$
$2.$ $T_r$ is always
$(A)$ an odd number
$(B)$ an even number
$(C)$ a prime number
$(D)$ a composite number
$3.$ Which one of the following is a correct statement?
$(A)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $5$
$(B)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $6$
$(C)$ $Q_1, Q_2, Q_3, \ldots$ are in $A.P.$ with common difference $11$
$(D)$ $Q_1 = Q_2 = Q_3 = \ldots$

An arithmetic progression is written in the following way. The sum of all the terms of the $10^{\text{th}}$ row is..........

Let $\alpha = \sum_{n=101}^{200} 2^n \sum_{k=101}^n \frac{1}{k !}$ and $b = \sum_{n=101}^{200} \frac{2^{201}-2^n}{n !}$. Then,$\frac{a}{b}$ is

Let the first term of a series be $T_1=6$ and its $r^{\text{th}}$ term $T_r=3T_{r-1}+6^r$ for $r=2, 3, \ldots, n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}(n^2-12n+39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$,then $n$ is equal to:

Consider an arithmetic series and a geometric series having four initial terms from the set $\{11, 8, 21, 16, 26, 32, 4\}$. If the last terms of these series are the maximum possible four-digit numbers,then the number of common terms in these two series is equal to .......