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Mathematics · 2023

JEE Main · 8 April 2023, Shift 1 · Q9

Let S_K=(1+2+…+K)/K and Σ_j=1^n S_j^2=n/A(B n^2+C n+D), where A, B, C, D ∈ N and A has least value. Then

Let $\displaystyle S_K=\frac{1+2+\ldots+K}{K}$ and $\displaystyle \sum_{j=1}^n S_j^2=\frac{n}{A}\left(B n^2+C n+D\right)$, where $\displaystyle A, B, C, D \in \mathbb{N}$ and $\displaystyle A$ has least value. Then
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.