Mathematics · 2019
JEE Main · 12 April 2019, Shift 1 · Q69
For x ∈ R, let [x] denote the greatest integer ≤ x, then the sum of the series [-1/3]+[-1/3-1/100]+[-1/3-2/100]+⋯+[-1/3-99/100] is:
For $\displaystyle x \in \mathbf{R}$, let $\displaystyle [x]$ denote the greatest integer $\displaystyle \leq x$, then the sum of the series
$$\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\cdots+\left[-\frac{1}{3}-\frac{99}{100}\right]
$$
is :
Official answer
(4)
$\displaystyle -133$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.