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
\(\begin{aligned} t_2 & =-1 \\ t_4 & =\left(\frac{4-3}{4-1}\right) \times t_{4-2} \\
& =\frac{1}{3} \times t_2=\frac{1}{3} \times(-1)=-\frac{1}{3} \\ t_6 &
=\left(\frac{6-3}{6-1}\right) \times t_{6-2} \\ & =\frac{3}{5} \times t_4=\frac{3}{5}
\times-\frac{1}{3}=-\frac{1}{5} \\ t_8 & =\left(\frac{8-3}{8-1}\right) \times t_{8-2} \\ &
=\frac{5}{7} \times t_6=\frac{5}{7} \times-\frac{1}{5}=-\frac{1}{7}\end{aligned}\)
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"
Choice C is the correct answer.
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