We are given the sum: \[ a_1 + a_2 + \dots + a_{2024} = 2233 \] In an Arithmetic Progression (A.P.), the sum of terms equidistant from the ends is equal, so: \[ a_1 + a_{2024} = a_2 + a_{2023} = \dots = a_{1012} + a_{1013} \] Thus, the number of pairs is: \[ 203 \quad \text{pairs of the form} \quad (a_1 + a_{2024}) \] Hence, we calculate: \[ S_{2024} = \frac{2024}{2} (a_1 + a_{2024}) = 2233 \] Now using the sum of A.P. formula, we get: \[ S = 2024 \times 11 \] \(\text{Therefore, the final sum is:}\) \[ \boxed{11132} \]
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)