If $a, b, c$ are in $A.P.$,then $\frac{a}{bc}, \frac{1}{c}, \frac{2}{b}$ are in

Let $a_1, a_2, a_3, \ldots$ be an $A$.$P$. If $a_7 = 3$,the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero,then $n! - 4 a_{n(n+2)}$ 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 $x_1, x_2, x_3, x_4$ be in a geometric progression. If $2, 7, 9, 5$ are subtracted respectively from $x_1, x_2, x_3, x_4$,then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24}(x_1 x_2 x_3 x_4)$ is:

Let $2^{\text{nd}}$,$8^{\text{th}}$,and $44^{\text{th}}$ terms of a non-constant $A.P.$ be respectively the $1^{\text{st}}$,$2^{\text{nd}}$,and $3^{\text{rd}}$ terms of a $G.P.$ If the first term of the $A.P.$ is $1$,then the sum of the first $20$ terms is equal to:

Consider the sequence $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{1}=1, a_{2}=2$ and $a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$ for $n=1, 2, 3, \ldots$. If $\left(\frac{a_{1}+\frac{1}{a_{2}}}{a_{3}}\right) \cdot\left(\frac{a_{2}+\frac{1}{a_{3}}}{a_{4}}\right) \cdot\left(\frac{a_{3}+\frac{1}{a_{4}}}{a_{5}}\right) \cdots\left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=2^{\alpha}\left({}^{61}C_{31}\right)$,then $\alpha$ is equal to.