$\text{ Let }{s}_1\text{ be the sum of first }2\mathrm{n}\text{ terms of an arithmetic progression. Let }{s}_2\text{ be the sum of first }$

$\text{ 4n terms of the same arithmetic progression. If }\left(s_2-s_1\right)\text{ is }1000,\text{ then the sum of the first }$$6n terms of the arithmatic progression is equal to$

$\begin{array}{l}\text{Suppose the given arithmetic progresion is having first term }a,\text{\hspace{0.17em} and the common difference is\hspace{0.17em}}d\\ \text{Hence, }\\ S_1=S_{2n}\\ =\frac{2n}{2}\left(2a+\left(2n-1\right)d\right)\\ =n\left(2a+2nd-d\right)\\ S_2=S_{4n}\\ =\frac{4n}{2}\left(2a+\left(4n-1\right)d\right)\\ =2n\left(2a+4nd-d\right)\\ {\text{Given S}}_2-S_1=1000\\ n\left(4a+8nd-2d-2a-2nd+d\right)=1000\\ n\left(2a+6nd-d\right)=1000\\ n\left(2a+\left(6n-1\right)d\right)=1000\\ \frac{6n}{2}\left(2a+\left(6n-1\right)d\right)=3\left(1000\right)\\ \text{Sum of the first 6n term is 3000}\end{array}$