Let $S_{1}$ be the sum of first $2 n$ terms of an arithmetic progression. Let $S_{2}$ be the sum of first 4n 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]
$\Rightarrow S_{2}-S_{1}=\frac{4 n}{2}[2 a+(4 n-1) d]-\frac{2 n}{2}[2 a+(2 n-$
1)d]
$=4 \mathrm{an}+(4 \mathrm{n}-1) 2 \mathrm{nd}-2 \mathrm{na}-(2 \mathrm{n}-1) \mathrm{dn}$
$=2 \mathrm{na}+\mathrm{nd}[8 \mathrm{n}-2-2 \mathrm{n}+1]$
$\Rightarrow 2 \mathrm{na}+\mathrm{nd}[6 \mathrm{n}-1]=1000$
$2 \mathrm{a}+(6 \mathrm{n}-1) \mathrm{d}=\frac{1000}{\mathrm{n}}$
Now, $S_{6 n}=\frac{6 n}{2}[2 a+(6 n-1) d]$
$=3 \mathrm{n} \cdot \frac{1000}{\mathrm{n}}=3000$
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