JEE PYQ: Functions Question 30
Question 30 - 2019 (12 Apr Shift 1)
For $x \in \mathbb{R}$, let $[x]$ denote the greatest integer $\le x$, 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:
(1) $-153$
(2) $-133$
(3) $-131$
(4) $-135$
Type: MCQ
Show Answer
Answer: (2) $-133$
Solution
$= -100 - \left\lfloor\frac{100}{3}\right\rfloor = -100 - 33 = -133$
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