CAT 2025 Slot 1 QA Q4 Progressions & Series

Question 4 of the Quantitative Ability (QA) section, CAT 2025 Slot 1, with CAT's official answer and a worked solution.

In the set of consecutive odd numbers \(\{1,3,5, \ldots, 57\}\), there is a number \(k\) such that the sum of all the elements less than \(k\) is equal to the sum of all the elements greater than \(k\). Then, \(k\) equals
  1. A
    41
  2. B
    39
  3. C
    37
  4. D
    43

Answer: (A) 41 Progressions & Series MCQ

Solution

  1. The sum of the first \(m\) odd numbers is \(m^2\). Let \(k = 2m + 1\), so the elements below \(k\) sum to \(m^2\).
  2. The set has 29 terms with total \(29^2 = 841\). The elements above \(k\) sum to \(841 - (m+1)^2\).
  3. \(m^2 = 841 - (m+1)^2 \Rightarrow m^2 + m = 420 \Rightarrow m = 20\).
  4. \(k = 41\).

Shortcut. Test options: for \(k = 41\), the sum below is \(20^2 = 400\) and the sum above is \(841 - 441 = 400\).

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