NinetynineCAT 2024 Slot 3 QA Q17 Progressions & Series
Question 17 of the Quantitative Ability (QA) section, CAT 2024 Slot 3, with CAT's official answer and a worked solution.
Consider the sequence \(t_1=1, t_2=-1\) and \(t_n=\left(\frac{n-3}{n-1}\right) t_{n-2}\) for \(n \geq 3\). Then, the value of the sum \(\frac{1}{t_2}+\frac{1}{t_4}+\frac{1}{t_6}+\cdots+\frac{1}{t_{2022}}+\frac{1}{t_{2024}}\), is
- A-1023132
- B-1022121
- C-1024144
- D-1026169
Answer: (C) -1024144 Progressions & Series MCQ
Solution
- \(t_4 = \frac13 t_2 = -\frac13\), \(t_6 = \frac35 t_4 = -\frac15\); the product telescopes to \(t_{2k} = -\frac{1}{2k-1}\).
- So \(\frac{1}{t_{2k}} = -(2k-1)\), and the sum for \(k = 1\) to 1012 is \(-(1 + 3 + \cdots + 2023) = -1012^2 = -1024144\).
Shortcut. Sum of first \(n\) odd numbers is \(n^2\); only C is \(-1012^2\).