CAT 2024 Slot 3 QA Question & Solution
Question
Consider the sequence $t_1 = 1, t_2 = -1$ and $t_n = \left(\cfrac{n - 3}{n - 1}\right)t_{n - 2}$ for $n \geq 3$. Then, the value of the sum $\cfrac{1}{t_2} + \cfrac{1}{t_4} + \cfrac{1}{t_6} + ....... +\cfrac{1}{t_{2022}} + \cfrac{1}{t_{2024}}$, is
Options
Solution
Finding the terms in the sequence, we see that $t_3=0$, $t_4=-\dfrac{1}{3}$, $t_5=0$
We would notice that all the odd terms are 0, and we are also asked the sum of only even terms, so we do not need to consider those
$t_6=-\dfrac{1}{5}$
We see that the even terms are in an HP: $-1,\ -\dfrac{1}{3},\ -\dfrac{1}{5},\ -\dfrac{1}{7},\ ...$
The sum we are asked is the inverse of these terms, that is: -1, -3, -5, -7, up to 1012 terms
The sum of this AP would be $\dfrac{\left[-\left(2\times\ 1\right)+\left(1012-1\right)\left(-2\right)\right]}{2}\times\ 1012$
Which is equal to $-1012\times\ 1012\ =\ -1024144$
Therefore, Option A is the correct answer.