Let $A=\left\{n \in N \mid n^{2} \leq n+10,000\right\}, B=\{3 k+1 \mid k \in N\}$ and $C=\{2 k \mid k \in N\}$, then the sum of all the elements of the set $A \cap(B-C)$ is equal to ______

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Sets Relations and Functions

Let

A = {n \in N ∣ n^{2} \leq n + 10, 000}, B = {3 k + 1 ∣ k \in N}

and

C = {2 k ∣ k \in N}

, then the sum of all the elements of the set

A \cap (B - C)

is equal to ______

Answer: 832

The condition

n^{2} \leq n + 10000

means

n^{2} - n - 10000 \leq 0

, so

n \leq \frac{1 + 40001}{2}

, and since

40001

exceeds

200

by less than

0.01

, this gives

n \leq 100

and

A = {1, 2, \dots, 100}

.
With

k

running over the natural numbers,

B = {4, 7, 10, 13, \dots}

and

C

is the set of even natural numbers, so

B - C

keeps exactly the odd members of

B

. Now

3 k + 1

is odd precisely when

k

is even, so

B - C = {7, 13, 19, 25, \dots}

the numbers of the form

6 m + 1

with

m \geq 1

. Those lying in

A

are

7, 13, \dots, 97

, an arithmetic progression of

16

terms, whose sum is

\frac{16}{2} (7 + 97) = 8 \times 104 = 832

So the sum of all the elements of

A \cap (B - C)

832

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer