Modern Math

Set Theory

Medium

Let A and B be two finite sets such that $n(A - B)$ , $n(A\cap B)$ , $n(B - A)$ are in an arithmetic progression. Here $n(X)$ denotes the number of elements in a finite set $X$ . If $n(A\cup B) = 18$ , then $n(A) + n(B)$ is ____

✅ Correct Option: 1

Let $n(A-B) = x$ , $n(A\cap B) = y$ , and $n(B-A) = z$ .

Since these three values form an arithmetic progression, we have:

$y - x = z - y$

This gives us: $2y = x+z$

We know that $n(A\cup B) = x + y + z = 18$

Substituting $2y = x+z$ into $x + y + z = 18$ :

$x + \dfrac{x+z}{2} + z = 18$

$\dfrac{3x}{2} + \dfrac{3z}{2} = 18$

$3(x+z) = 36$

$x + z = 12$

Since $2y = x+z = 12$ , we have $y = 6$

Using the set theory formulas:

$n(A) = n(A-B) + n(A\cap B) = x + y = x + 6$

$n(B) = n(B-A) + n(A\cap B) = z + y = z + 6$

Therefore:

$n(A) + n(B) = x + 6 + z + 6 = x + z + 12 = 12 + 12 = 24$

The answer is $24$ .

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