CAT 2021 Slot 3QA Question 17

Arithmetic ProgressionEasy

Consider a sequence of real numbers x1, x2, x3, … such that xn+1 = xn + n – 1 for all n ≥ 1. If x1 = -1 then x100

Answer & solution

  • 4850

  • B

    4849

  • C

    4950

  • D

    4949

Solution

Easy

Unroll the recurrence xn+1=xn+(n1)x_{n+1}=x_n+(n-1) by summing the increments from x1x_1 up to x100x_{100}. The increments form 0,1,2,,980,1,2,\dots,98, a simple arithmetic-series sum.

1

List the increments. Going from xnx_n to xn+1x_{n+1} adds n1n-1.

x2=x1+0x3=x2+1x4=x3+2, \begin{aligned} &x_2 = x_1 + 0\\ &x_3 = x_2 + 1\\ &x_4 = x_3 + 2,\ \dots \end{aligned}
2

Telescope up to x100x_{100}. The increments added are 0+1+2++980+1+2+\cdots+98.

x100=x1+(0+1+2++98) x100=x1+98992(sum of 0..98) x100=x1+4851\begin{aligned} &x_{100} = x_1 + (0+1+2+\cdots+98)\\ &\Rightarrow\ x_{100} = x_1 + \frac{98\cdot 99}{2}\quad\text{(sum of }0..98)\\ &\Rightarrow\ x_{100} = x_1 + 4851 \end{aligned}
3

Substitute x1=1x_1=-1.

x100=1+4851=4850\begin{aligned} &x_{100} = -1 + 4851 = 4850 \end{aligned}
x100=4850x_{100} = 4850
CAT 2021 Slot 3 QA Q17: Consider a sequence of real numbers x 1 , x 2 , x 3 , … such that x n+1 = x n + n – 1 for all n & — Solution | TheCATExam