Mathematics · 2023
JEE Main · 24 January 2023, Shift 2 · Q83
If (1^3+2^3+3^3+… up to n terms)/(1 · 3+2 · 5+3 · 7+… up to n terms)=9/5, then the value of n is
If $\displaystyle \frac{1^3+2^3+3^3+\ldots \text { up to } n \text { terms }}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \text { up to } n \text { terms }}=\frac{9}{5}$, then the value of $\displaystyle n$ is
Official answer
From NTA’s final answer key for this paper.
5
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.