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 $a_n = \frac{10^n}{n!}$ for $n = 1, 2, 3, \ldots$. The greatest value of $n$ for which $a_n$ is the greatest is:

If the sum of the first $40$ terms of the series $3+4+8+9+13+14+18+19+\ldots$ is $(102)m$,then $m$ is equal to:

Let $I(n) = n^n$ and $J(n) = 1 \times 3 \times 5 \times \ldots \times (2n - 1)$ for all $n > 1, n \in N$. Then:

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

Let $a, b, c > 1$. If $a^3, b^3, c^3$ are in $A.P.$ and $\log_a b, \log_c a, \log_b c$ are in $G.P.$,and the sum of the first $20$ terms of an $A.P.$ with first term $\frac{a+4b+c}{3}$ and common difference $\frac{a-8b+c}{10}$ is $-444$,then $abc$ is equal to: