CAT 2024 Question Paper | Quant Slot 3

CAT Previous Year Paper | CAT Quant Questions | Question 17

CAT 2024 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 17 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

  1. -1023132
  2. -1022121
  3. -1024144
  4. -1026169

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Explanatory Answer

\(\begin{aligned} t_2 & =-1 \\ t_4 & =\left(\frac{4-3}{4-1}\right) \times t_{4-2} \\ & =\frac{1}{3} \times t_2=\frac{1}{3} \times(-1)=-\frac{1}{3} \\ t_6 & =\left(\frac{6-3}{6-1}\right) \times t_{6-2} \\ & =\frac{3}{5} \times t_4=\frac{3}{5} \times-\frac{1}{3}=-\frac{1}{5} \\ t_8 & =\left(\frac{8-3}{8-1}\right) \times t_{8-2} \\ & =\frac{5}{7} \times t_6=\frac{5}{7} \times-\frac{1}{5}=-\frac{1}{7}\end{aligned}\)
Then,
\[ \begin{array}{l} \frac{1}{t_2}+\frac{1}{t_4}+\frac{1}{t_6}+\frac{1}{t_8}+\ldots+\frac{1}{t_{2024}}=(-1)+(-3)+(-5)+(-7)+\ldots+(2023) \\ =-(1+3+5+7+\ldots+2023) \\ =-(\text { Sum of first } 1012 \text { odd numbers }) \\ =-(1012)^2 \\ =-1024144 \end{array} \]


The question is "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"

Hence, the answer is '-1024144'

Choice C is the correct answer.

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