Let the sum of an infinite $G.P.$,whose first term is $a$ and the common ratio is $r$,be $5$. Let the sum of its first five terms be $\frac{98}{25}$. Then the sum of the first $21$ terms of an $A.P.$,whose first term is $10ar$,$n^{\text{th}}$ term is $a_n$ and the common difference is $10ar^2$,is equal to.

If $m$ is the $A.M.$ of two distinct real numbers $l$ and $n$ $(l, n > 1)$ and $G_1, G_2, G_3$ are three geometric means between $l$ and $n$,then $G_1^4 + 2G_2^4 + G_3^4$ equals:

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

Let $\{a_k\}$ and $\{b_k\}, k \in N$,be two $G$.$P$.s with common ratios $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+b_k, k \in N$. If $c_2=5$ and $c_3=13/4$,then $\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$ is equal to

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