Let a1, a2, …, a2024 be an Arithmetic Progression such that a1 + (a5 + a10 + a15 + … + a2020) + a2024 = 2233. Then a1 + a2 + a3 + … + a2024 is equal to________.

a1 + a5 + a10 + a15 + … + a2020 + a2024 = 2233In an A.P. the sum of terms equidistant from endsis equal.a1 + a2024 = a5 + a2020 = a10 + a2015 …..⇒203 pairs⇒203(a1 + a2024) = 2233Hence,S2024=20242a1+a2024= 1012 × 11= 11132

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Let a₁, a₂, …, a₂₀₂₄ be an Arithmetic Progression such that a₁ + (a₅ + a₁₀ + a₁₅ + … + a₂₀₂₀) + a₂₀₂₄ = 2233. Then a₁ + a₂ + a₃ + … + a₂₀₂₄ is equal to
________.

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answer is 11132.

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Detailed Solution

a₁ + a₅ + a₁₀ + a₁₅ + … + a₂₀₂₀ + a₂₀₂₄ = 2233
In an A.P. the sum of terms equidistant from ends
is equal.
a₁ + a₂₀₂₄ = a₅ + a₂₀₂₀ = a₁₀ + a₂₀₁₅ …..
$\Rightarrow$ 203 pairs
$\Rightarrow$ 203(a₁ + a₂₀₂₄) = 2233
Hence,
$S_{2024} = \frac{2024}{2} (a_{1} + a_{2024})$
= 1012 × 11
= 11132