CAT Exam 2021 Slot 3 Solved Paper

Given :
$\x_\n + 1 =\x_\n + \n- 1 $ and $\x_1 = -1$
Considering :
$\x_1 = -1$ (1)
$\x_2 = x_1+1-1 = \x_1 + 0$ (2)
$\x_3 = \x_2 + 2 - 1 =\x_2 + 1$ (3)
$\x_4 = \x_3 + 3 - 1 = \x_3 + 2$ (4)
$\x_100 = \x_99 + 98$ (100)
Adding the LHS and RHS for the hundred equations we have:
$(\x_1+\x_2+............\x100) = (-1+0+.........98) + (\x_1+\x_2+........\x_99)$
Subtracting this we have :
$(\x_1+...........\x_100) - (\x_1+............. \x_99) = {98.99/2} - 1.$
$\x_100 = 4851 - 1 = 4850$

Alternatively :
$\x_1 = -1$
$\x_2 = \x_1 + 1 -1 = \x_1 = -1 $
$\x_3 = \x_2 + 2 -1 = \x_2 + 1 = -1 +1 = 0 $
$x_4 = x_3 + 3 -1 = x_3 + 2 = 0 +2 = 2 $
$\x_5 = \x_4 + 4 -1 = \x_4 + 3 = 2 + 3= 5 $
...............
If we observe the series, it is a series that has a difference between the consecutive terms in an AP.
Such series are represented as $\t (\n) = \a + \bn + \cn^2$
We need to find t(100).
$ \t(1) = -1 $
$ \a + \b + \c = -1 $
$\t(2) = -1$
$ \a + 2\b + 4\c = -1$
$ \t(3) = 0 $
$\a + 3\b + 9\c = 0 $
Solving we get,
$\b + 3\c = 0 $
$ \b + 5\c = 1 $
$\c = 0.5$
$\b = -1.5 $
$ \a = 0$
Now,
$\t(100) = (-1.5) 100 + (0.5)100^2 = -150 + 5000 = 4850$