Let A- n-n2- n-10-000 - B- 3k-1-k - and C-2k-k- Then the sum of all the elements of the set A- B-C is equal to

N^2 n 10000≤0 n∈(1 √(40001)2, 1+√(40001)2) But n∈ℕ ∴ A={1,2,3,.......,100} and B C={3k+1|k∈even.} ⇒ B C={7,13,19,.......,97,..} ∴ A∩ (B C)={7,13,19,...,97} Sum ...

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Standard XII

Mathematics

Fundamental Laws of Logarithms

Let A= n∈ℕ|n2...

Question

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

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Solution

$n^{2} - n - 10000 \leq 0$
$n \in (\frac{1 - \sqrt{40001}}{2}, \frac{1 + \sqrt{40001}}{2})$
But $n \in N$
$∴ A = {1, 2, 3, . . . . . . ., 100}$
$and B - C = {3 k + 1 | k \in even}$
$\Rightarrow B - C = {7, 13, 19, . . . . . . ., 97, . .}$
$∴ A \cap (B - C) = {7, 13, 19, . . ., 97}$
Sum of all elements $= \frac{16}{2} [7 + 97] = 832$

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Let A={n∈N∣∣n2≤n+10,000}, B={3k+1|k∈N} and C={2k|k∈N}. Then the sum of all the elements of the set A∩(B−C) is equal to

n2−n−10000≤0 n∈(1−√400012,1+√400012) But n∈N ∴A={1,2,3,.......,100} and B−C={3k+1|k∈even} ⇒B−C={7,13,19,.......,97,..} ∴A∩(B−C)={7,13,19,...,97} Sum of all elements =162[7+97]=832

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