Let S_{1} be the sum of first 2 n terms of an | vedclass

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Question

$S_{2 n}=\frac{2 n}{2}[2 a+(2 n-1) d], S_{4 n}=\frac{4 n}{2}[2 a+(4 n-1)d]$

$\Rightarrow S _{2}- S _{1}=\frac{4 n }{2}[2 a +(4 n -1) d ]-\frac{2 n

Vedclass · Accepted Answer

$3000$

Vedclass · Answer

$S_{2 n}=\frac{2 n}{2}[2 a+(2 n-1) d], S_{4 n}=\frac{4 n}{2}[2 a+(4 n-1)d]$

$\Rightarrow S _{2}- S _{1}=\frac{4 n }{2}[2 a +(4 n -1) d ]-\frac{2 n }{2}[2 a +(2 n -1)d]$

$=4 a n+(4 n-1) 2 n d-2 n a-(2 n-1) d n$

$=2 n a+n d[8 n-2-2 n+1]$

$\Rightarrow 2 n a+n d[6 n-1]=1000$

$2 a+(6 n-1) d=\frac{1000}{n}$

Now, $S_{6 n}=\frac{6 n}{2}[2 a+(6 n-1) d]$

$=3 n \cdot \frac{1000}{ n }=3000$

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:

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