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  • Help me solve this - Sets, Relations and Functions - JEE Main-8

 

For x\equiv R, let \left [ x \right ] denote the greatest integer \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 : 


 

  • Option 1)

    -131

  • Option 2)

    -133

  • Option 3)

    -135

  • Option 4)

    -153

 

Answers (1)

\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 ]

\underbrace{\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+.........+\left[-\frac{1}{3}-\frac{66}{100}\right]}+\underbrace{\left[-\frac{1}{3}-\frac{67}{100}\right]+.........+\left[-\frac{1}{3}-\frac{99}{100}\right]}

                                     (-1)\times 64                                                                        (-2)\times 33

=-67-66

=-133


Option 1)

-131

Option 2)

-133

Option 3)

-135

Option 4)

-153

Posted by

Vakul

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