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CAT 2021 Slot 1 — QA Question 22

Miscellaneous ProgressionsEasy

The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15th group is equal to

Answer & solution

  • A

    4941

  • 6119

  • C

    7471

  • D

    6090

Solution

Easy

The group sizes are 1,3,5,…1,3,5,\dots (odd numbers), so the count of numbers used up to and including the nn-th group is the perfect square n2n^2. That pins down exactly which natural numbers sit in the 1515th group, after which sum them as an arithmetic progression.

1

Group sizes and running totals. The nn-th group has (2n−1)(2n-1) numbers, and the total used through group nn is

1+3+5+⋯+(2n−1)=n2\begin{aligned} &1+3+5+\cdots+(2n-1)=n^2 \end{aligned}
2

Locate the 15th group. Numbers through group 1414 end at 142=19614^2=196; through group 1515 end at 152=22515^2=225.

15th group=197, 198, …, 225⇒ it has 2(15)−1=29 numbers\begin{aligned} &\text{15th group}=197,\,198,\,\dots,\,225\\ &\Rightarrow\ \text{it has }2(15)-1=29\ \text{numbers} \end{aligned}
3

Sum the arithmetic progression from 197197 to 225225.

Sum=count2 (first+last)=292 (197+225)⇒ =292×422=29×211=6119\begin{aligned} &\text{Sum}=\frac{\text{count}}{2}\,(\text{first}+\text{last})=\frac{29}{2}\,(197+225)\\ &\Rightarrow\ =\frac{29}{2}\times 422=29\times 211=6119 \end{aligned}
6119 (option b)6119\ \text{(option b)}

The middle (median) term of the 1515th group is 152−15+1=21115^2-15+1=211, the central element; with 2929 equally spaced terms the sum is 29×211=611929\times211=6119.

CAT 2021 Slot 1 QA Q22: The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, th — Solution | TheCATExam