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A=200,201,202.........,2000
Here, it forms an AP with common difference 1
Let the number of terms in the set be n
[tex] T_{n} = a+(n-1)d \\\\ 2000= 200+(n-1) \\\\ 200+n-1= 2000 \\\\ => n=2000-199= 1801[/tex]
Hence,n(A)=1801