If $a, b, c$ are in $G.P.$ and $x, y$ are the arithmetic means between $a, b$ and $b, c$ respectively,then $\frac{a}{x} + \frac{c}{y}$ is equal to

Let $a_1=b_1=1$ and $a_n=a_{n-1}+(n-1)$,$b_n=b_{n-1}+a_{n-1}$,$\forall n \geq 2$. If $S =\sum \limits_{n=1}^{10} \frac{b_n}{2^n}$ and $T =\sum \limits_{n=1}^8 \frac{n}{2^{n-1}}$,then $2^7(2S - T)$ is equal to $........$.

Three numbers form a $G.P.$ If the $3^{rd}$ term is decreased by $64$,the three numbers thus obtained constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$,a $G.P.$ is formed again. Find the numbers.

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

If $a$ is the arithmetic mean of $b$ and $c$ and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $