Let a_1, a_2, ldots, a_{2024} be an Arithmeti | vedclass

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(D) Let the given sum be $S = a_1 + (a_5 + a_{10} + \ldots + a_{2020}) + a_{2024} = 2233$. In an $A.P.$,the sum of terms equidistant from the beginnin

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$11132$

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(D) Let the given sum be $S = a_1 + (a_5 + a_{10} + \ldots + a_{2020}) + a_{2024} = 2233$. In an $A.P.$,the sum of terms equidistant from the beginning and end is constant,i.e.,$a_k + a_{n-k+1} = a_1 + a_n$. Here,$n = 2024$. The terms in the bracket are $a_{5k}$ for $k=1$ to $404$. Note that $a_5 + a_{2020} = a_1 + a_{2024}$,$a_{10} + a_{2015} = a_1 + a_{2024}$,and so on. There are $404$ terms in the sequence $5, 10, \ldots, 2020$. Pairing these terms,we have $202$ pairs,each equal to $(a_1 + a_{2024})$. Including the outer terms $a_1$ and $a_{2024}$,the total sum is $203(a_1 + a_{2024}) = 2233$. Thus,$(a_1 + a_{2024}) = \frac{2233}{203} = 11$. The sum of the $A.P.$ is $S_{2024} = \frac{2024}{2}(a_1 + a_{2024}) = 1012 	imes 11 = 11132$.

Let $a_1, a_2, \ldots, a_{2024}$ be an Arithmetic Progression such that $a_1 + (a_5 + a_{10} + a_{15} + \ldots + a_{2020}) + a_{2024} = 2233$. Then $a_1 + a_2 + a_3 + \ldots + a_{2024}$ is equal to . . . . . .

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