Question

Let \( a_1, a_2, \dots, a_{2024} \) be an Arithmetic Progression such that \[ a_1 + (a_1 + a_0 + a_1 + a_2 + \cdots + a_{2020} + a_{2024}) = 2233. \quad \text{Then} \quad a_1 + a_2 + a_3 + \dots + a_{2022} \] is equal to ____ :

Hint:

When dealing with Arithmetic Progression, utilize the formula for the sum of terms and the property of sums of equidistant terms for efficient calculations.

Answer 1

Correct Answer: 11132

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: \[ S = 2024 \times 11 \] \(\text{Therefore, the final sum is:}\) \[ \boxed{11132} \]