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 $........$.

Let $S$ be the sum of the first $9$ terms of the series: $(x+ka) + (x^2+(k+2)a) + (x^3+(k+4)a) + (x^4+(k+6)a) + \ldots$ where $a \neq 0$ and $x \neq 1$. If $S = \frac{x^{10}-x+45a(x-1)}{x-1}$,then $k$ is equal to:

Let $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \ldots$ up to $n$ terms. If the sum of the first six terms of an $A.P.$ with first term $-p$ and common difference $p$ is $\sqrt{2026 S_{2025}}$,then the absolute difference between the $20^{\text{th}}$ and $15^{\text{th}}$ terms of the $A.P.$ is

Let $\langle a_n \rangle$ be a sequence such that $a_0 = 0, a_1 = \frac{1}{2}$ and $2a_{n+2} = 5a_{n+1} - 3a_n$ for $n = 0, 1, 2, 3, \ldots$. Then $\sum_{k=1}^{100} a_k$ is equal to:

If $\log_x y, \log_z x, \log_y z$ are in $G.P.$,$xyz = 64$,and $x^3, y^3, z^3$ are in $A.P.$,then

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: