MCQMediumJEE Main 2023 · 8 April, Shift 1Sum of Series

Mathematics Question from JEE Main 2023 · 8 April, Shift 1

Let SK=1+2++KKS_K = \frac{1 + 2 + \dots + K}{K} and j=1nSj2=nA(Bn2+Cn+D)\sum_{j=1}^{n} S_j^2 = \frac{n}{A}\left(Bn^2 + Cn + D\right), where A,B,C,DNA, B, C, D \in \mathbb{N} and AA has the least value. Then:

  • A

    A+BA + B is divisible by DD

  • B

    A+B=5(DC)A + B = 5(D - C)

  • C

    A+C+DA + C + D is not divisible by BB

  • D

    A+B+DA + B + D is divisible by 55

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Practice more Sum of Series questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Related questions