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CAT 2020 Slot 3 — QA Question 6

Arithmetic ProgressionEasy

If x₁ = - 1 and xm = xm+1 + (m + 1) for every positive integer m, then x100 equals

Answer & solution

Correct answer: -5050

  • A

    -5151

  • B

    -5150

  • -5050

  • D

    -5051

Solution

Easy

Rearrange the recurrence to express xm+1x_{m+1} in terms of xmx_m, unfold a few terms to spot the pattern, then sum.

1

Rearrange. Given xm=xm+1+(m+1)x_m=x_{m+1}+(m+1).

xm+1=xm−(m+1)\begin{aligned} &x_{m+1}=x_m-(m+1) \end{aligned}
2

Unfold from x1=−1x_1=-1.

x2=x1−2=−1−2(m=1)x3=x2−3=−1−2−3(m=2)x4=x3−4=−1−2−3−4(m=3)\begin{aligned} &x_2=x_1-2=-1-2 \quad\text{(}m=1)\\ &x_3=x_2-3=-1-2-3 \quad\text{(}m=2)\\ &x_4=x_3-4=-1-2-3-4 \quad\text{(}m=3) \end{aligned}
3

General term and sum. The pattern gives x100=−(1+2+3+⋯+100)x_{100}=-(1+2+3+\cdots+100), the negative of the sum of the first 100100 natural numbers.

x100=−100×1012⇒ x100=−5050\begin{aligned} &x_{100}=-\frac{100\times101}{2}\\ &\Rightarrow\ x_{100}=-5050 \end{aligned}
x100=−5050x_{100}=-5050

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CAT 2020 Slot 3 QA Q6 — Arithmetic Progression solution | TheCATExam