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:

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 the integers from $1$ to $2021$ are written as a single integer like $123 \dots 91011 \dots 20202021$,then the $2021^{st}$ digit (counted from the left) in the resulting number is

Let $\{a_{n}\}_{n=0}^{\infty}$ be a sequence such that $a_{0}=0, a_{1}=0$ and $a_{n+2}=3a_{n+1}-2a_{n}+1$ for all $n \geq 0$. Then the value of $a_{25}a_{23}-2a_{25}a_{22}-2a_{23}a_{24}+4a_{22}a_{24}$ is equal to:

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