CAT 2020 Quant Question [Slot 3] with Solution 05

Question

If \({x_1} =  - 1\) and \({x_m} = {x_{m + 1}} + (m + 1)\) for every positive integer $m,$ then \({x_{100}}\) equals

  1. -5151
  2. -5150
  3. -5051
  4. -5050
Option: 4
Solution:

\({x_m} + 1 = {x_m} - (m + 1)\)

\({x_2} = {x_1} - 2 =  - 1 - 2 =  - 3\)

\({x_3} = {x_2} - 3 =  - 1 - 2 - 3 =  - 6\)

Similarly,

\({x_n} =  - (1 + 2 + 3 +  \ldots  + n) =  - \frac{{n(n + 1)}}{2}\)

Hence \({x_{100}} =  - \frac{{100(101)}}{2} =  - 5050\)

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