1. Let the arithmetic progression have first term $a$ and common difference $d$.

2. Sum of first $m$ terms of an AP is $S_m = \frac{m}{2}\big[2a+(m-1)d\big].$

So, $S_{2n} = \frac{2n}{2}\big[2a+(2n-1)d\big] = n\big[2a+(2n-1)d\big].$

And, $S_{4n} = \frac{4n}{2}\big[2a+(4n-1)d\big] = 2n\big[2a+(4n-1)d\big].$

3. Given: $S_{4n}-S_{2n}=1000.$

Now, $S_{4n}-S_{2n}$ represents the sum of terms from $(2n+1)$-th to $4n$-th, i.e. exactly $2n$ terms.

Also, $S_{6n}-S_{4n}$ represents the sum of terms from $(4n+1)$-th to $6n$-th, again $2n$ terms.

Since these two blocks each contain $2n$ consecutive terms of an AP, their sums are not generally equal. So we compute directly.

4. Write the sums explicitly: $S_{2n}=n[2a+(2n-1)d],$ $S_{4n}=2n[2a+(4n-1)d],$ $S_{6n}=3n[2a+(6n-1)d].$

Now compute: $S_{4n}-S_{2n} = 2n[2a+(4n-1)d]-n[2a+(2n-1)d].$

Factor $n$: $= n\left(4a+2(4n-1)d-2a-(2n-1)d\right)$ $= n\left(2a+(8n-2-2n+1)d\right)$ $= n\left(2a+(6n-1)d\right).$

Given this equals $1000$, so $n\big(2a+(6n-1)d\big)=1000.$

5. Now find $S_{6n}$: $S_{6n}=3n[2a+(6n-1)d].$

Using the relation above, $S_{6n}=3\cdot 1000 = 3000.$

6. Therefore the correct option is $\boxed{3000}.$