Q.

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 …..
203 pairs
203(a1 + a2024) = 2233
Hence,
S2024=20242a1+a2024
= 1012 × 11
= 11132

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