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

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(D) Let the first term be $a$ and the common difference be $d$.$S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$$S_{4n} = \frac{4n}{2}[2a + (4n-

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$3000$

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(D) Let the first term be $a$ and the common difference be $d$.$S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$$S_{4n} = \frac{4n}{2}[2a + (4n-1)d] = 2n[2a + (4n-1)d]$Given $S_{2} - S_{1} = 1000$,where $S_{1} = S_{2n}$ and $S_{2} = S_{4n}$:$2n[2a + (4n-1)d] - n[2a + (2n-1)d] = 1000$$n[4a + 2(4n-1)d - 2a - (2n-1)d] = 1000$$n[2a + (8n - 2 - 2n + 1)d] = 1000$$n[2a + (6n - 1)d] = 1000$$2a + (6n - 1)d = \frac{1000}{n}$Now,the sum of the first $6n$ terms is $S_{6n} = \frac{6n}{2}[2a + (6n-1)d]$$S_{6n} = 3n 	imes \frac{1000}{n} = 3000$

Let $S_{1}$ be the sum of the first $2n$ terms of an arithmetic progression. Let $S_{2}$ be the sum of the first $4n$ terms of the same arithmetic progression. If $(S_{2} - S_{1})$ is $1000$,then the sum of the first $6n$ terms of the arithmetic progression is equal to:

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