Let $S_{1}$ be the sum of first $2 \mathrm{n}$ terms of an arithmetic progression. Let $S_{2}$ be the sum of first $4 \mathrm{n}$ terms of the same arithmetic progression. If $\left(S_{2}-S_{1}\right)$ is 1000 , then the sum of the first 6 n terms of the arithmetic progression is equal to:
Correct Option: , 4
$S_{2 n}=\frac{2 n}{2}[2 a+(2 n-1) d], S_{4 n}=\frac{4 n}{2}[2 a+(4 n-$1)d]
$\Rightarrow \mathrm{S}_{2}-\mathrm{S}_{1}=\frac{4 \mathrm{n}}{2}[2 \mathrm{a}+(4 \mathrm{n}-1) \mathrm{d}]-\frac{2 \mathrm{n}}{2}[2 \mathrm{a}+(2 \mathrm{n}-$
1)d] $=4 \mathrm{an}+(4 \mathrm{n}-1) 2 \mathrm{nd}-2 \mathrm{na}-(2 \mathrm{n}-1) \mathrm{dn}$
$=2 n a+n d[8 n-2-2 n+1]$
$\Rightarrow 2 \mathrm{na}+2 \mathrm{n}[6 \mathrm{n}-1]=1000$
$2 a+(6 n-1) d=\frac{1000}{n}$
Now, $S_{6 n}=\frac{6 n}{2}[2 a+(6 n-1) d]$
$=3 \mathrm{n} \cdot \frac{1000}{\mathrm{n}}=3000$