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JEE Mains · Maths · STD 11 - 8. sequence and series

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

  1. A 11132
  2. B 11134
  3. C 11136
  4. D 11138
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Answer & Solution

Correct Answer

(A) 11132

Step-by-step Solution

Detailed explanation

As \(a_1+a_5+a_{10}+\ldots+a_{2020}+a_{2024}=2233\) ...(1) We know in arithmetic progression. Sum of terms equidistant from ends is equal \(\therefore\) from (1) \(\underbrace{a_1+a_{2024}=a_5+a_{2020}=a_{10}+a_{2015}=\ldots}_{203 \text { pairs }}\)…
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