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 __________

Given, $$n(A - B), n(A \cap B), n(B - A)$$ are in A.P.

Now, $$n\left(A-B\right)=n\left(A\right)-n\left(A∩B\right)$$

$$n\left(B-A\right)=n\left(B\right)-n\left(A∩B\right)$$

Now, adding these two equations, $$n\left(A-B\right)+n\left(B-A\right)=n\left(A\right)+n\left(B\right)-2\cdot n\left(A∩B\right)$$--->(1)

Since, $$n(A - B), n(A \cap B), n(B - A)$$ are in A.P.

So, $$n\left(A-B\right)+n\left(B-A\right)=2\cdot n\left(A∩B\right)$$---->(2)

From (1) and (2), $$n\left(A\right)+n\left(B\right)-2\cdot n\left(A∩B\right)=2\cdot n\left(A∩B\right)$$

So, $$n\left(A\right)+n\left(B\right)=4\cdot n\left(A∩B\right)$$ ------>(3)

Now we know, $$n\ \left(A∪B\right)=n\left(A\right)+n\left(B\right)-n\left(A∩B\right)$$

or, $$18=4n\left(A∩B\right)-n\left(A∩B\right)=3.n\left(A∩B\right)$$  (From (3))

So, $$n\left(A∩B\right)=\dfrac{18}{3}=6$$

From (3), $$n\left(A\right)+n\left(B\right)=4\times\ 6=24$$