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Question 210012
JEE Mathematics · Hard

Sequences and Series: AP sum MCQ

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AP sumHardQuestion 210012

Question

Find the sum of all integers between 1 and 100 which are divisible by 3 or 5:
  1. A
    2317
  2. B
    2418
    Correct
  3. C
    2317+316
  4. D
    2633

Correct answer

2418

Explanation

Sum(÷3) + Sum(÷5) − Sum(÷15). Multiples of 3 in 1–100: 3,6,...,99, n=33, S=33×51=1683. Multiples of 5: 5,10,...,100, n=20, S=20×52.5=1050. Multiples of 15: 15,30,...,90, n=6, S=6×52.5=315. Total = 1683+1050−315 = 2418.

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