To solve the problem, we first need to understand the sequence \( S_n = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots \). This is a series where each term can be expressed as \(\frac{1}{n(n+1)}\). The terms of this series can be rewritten as a telescoping series: \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \] Thus, the sum of the series up to \( S_n \) is: \[ S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} \] Therefore, \( S_{2025} = 1 - \frac{1}{2026} \). Next, we address the A.P. problem. We are given that the sum of the first six terms of an A.P. with first term \(-p\) and common difference \(p\) is \(\sqrt{2026 S_{2025}}\). The sum of the first six terms (\(S_6\)) of an A.P. is: \[ S_6 = \frac{6}{2} \times [2(-p) + 5p] = 3(-2p + 5p) = 9p \] According to the problem: \[ 9p = \sqrt{2026 \times \left(1 - \frac{1}{2026}\right)} = \sqrt{2026 - 1} = \sqrt{2025} = 45 \] Thus, \(9p = 45\) which gives us \(p = 5\). We are required to find the absolute difference between the 20th and 15th terms of the A.P.: 20th term = \(-p + 19p = 19p - p = 18p\) 15th term = \(-p + 14p = 14p - p = 13p\) The absolute difference is: \[ |18p - 13p| = |5p| = 5 \times 5 = 25 \] Therefore, the absolute difference between the 20th and 15th terms of the A.P. is 25.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to:
\[ 5m \sum_{r=m}^{2m} T_r \text{ is equal to:} \]
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Two vessels A and B are connected via stopcock. Vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?
Choose the correct nuclear process from the below options:
\( [ p : \text{proton}, n : \text{neutron}, e^- : \text{electron}, e^+ : \text{positron}, \nu : \text{neutrino}, \bar{\nu} : \text{antineutrino} ] \)