Question ID: #578
Let $a_{1},a_{2},…,a_{2024}$ be an Arithmetic Progression such that $a_{1}+(a_{5}+a_{10}+a_{15}+…+a_{2020})+a_{2024} = 2233$. Then $a_{1}+a_{2}+a_{3}+…+a_{2024}$ is equal to
Solution:
In an A.P., the sum of terms equidistant from the beginning and end is constant ($a_1 + a_{2024} = a_5 + a_{2020} = \dots$).
First, we count the terms in the given sum.
The indices in the middle ($5, 10, \dots, 2020$) are multiples of 5.
Since $2020 = 5 \times 404$, there are 404 terms in the middle.
Adding $a_1$ and $a_{2024}$, the total number of terms is $1 + 404 + 1 = 406$.
We can group these 406 terms into pairs of $(a_1 + a_{2024})$.
Total pairs = $\frac{406}{2} = 203$.
The equation becomes:
$$ 203(a_1 + a_{2024}) = 2233 $$
$$ a_1 + a_{2024} = \frac{2233}{203} = 11 $$
Now, sum of first 2024 terms:
$$ S_{2024} = \frac{2024}{2}(a_1 + a_{2024}) $$
$$ S_{2024} = 1012 \times 11 $$
$$ S_{2024} = 11132 $$
Ans. (11132)
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