CAT 2021 Slot 3 QA Q16 Progressions & Series

Question 16 of the Quantitative Ability (QA) section, CAT 2021 Slot 3, with CAT's official answer and a worked solution.

Consider a sequence of real numbers \(x_{1}, x_{2}, x_{3}, \ldots\) such that \(x_{n+1}=x_{n}+n-1\) for all \(n \geq 1 .\) If \(x_{1}=-1\) then \(x_{100}\) is equal to
  1. A
    4949
  2. B
    4849
  3. C
    4850
  4. D
    4950

Answer: (C) 4850 Progressions & Series MCQ

Solution

  1. \(x_{100} = x_1 + \sum_{n=1}^{99}(n - 1) = -1 + (0 + 1 + \dots + 98)\).
  2. \(0 + 1 + \dots + 98 = \frac{98 \times 99}{2} = 4851\).
  3. \(x_{100} = 4850\).

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