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 product of three consecutive terms of a $G.P.$ is $512$. If $4$ is added to each of the first and the second of these terms,the three terms now form an $A.P.$ Then the sum of the original three terms of the given $G.P.$ is

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 $11$
$(D)$ $Q_1 = Q_2 = Q_3 = \ldots$

The sum of the first $n$ terms of a sequence is given by $S_n = 3n^2 + 4n + 15$. If $T_r$ is the $r^{th}$ term of the sequence,then $T_3 - T_1$ is equal to

Let $x_{1}, x_{2}, x_{3}, \ldots, x_{20}$ be in a geometric progression with $x_{1} = 3$ and common ratio $r = \frac{1}{2}$. $A$ new data set is constructed by replacing each $x_{i}$ with $(x_{i} - i)^{2}$. If $\bar{x}$ is the mean of the new data,then the greatest integer less than or equal to $\bar{x}$ is $.....$

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