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Mathematics · 2018

JEE Main · 16 April 2018, Shift 1 · Q68

Let 1/x_1, 1/x_2, …, 1/(x_n)(x_i ≠ 0. for.i=1,2, …, n ) be in A.P. such that x_1=4 and x_21=20. If n is the least positive integer for which x_n >50,…

Let $\displaystyle \frac{1}{x_1}, \frac{1}{x_2}, \ldots, \frac{1}{x_{\mathrm{n}}}\left(x_i \neq 0\right.$ for $\displaystyle \left.i=1,2, \ldots, \mathrm{n}\right)$ be in A.P. such that $\displaystyle x_1=4$ and $\displaystyle x_{21}=20$. If n is the least positive integer for which $\displaystyle x_{\mathrm{n}}>50$, then $\displaystyle \sum_{i=1}^{\mathrm{n}}\left(\frac{1}{x_i}\right)$ is equal to :
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JEE Main 2018 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.