Let $S_1$ be the sum of first $2n$ terms of an arithmetic progression. Let $S_2$ be the sum of 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:

Question

Aicharya · Accepted Answer

S 2 n = 2 n 2 2 a + 2 n - 1 d , &#160; S 4 n = 4 n 2 2 a + 4 n - 1 d &#8658; S 2 - S 1 = 4 n 2 [ 2 a + ( 4 n - 1 ) d ] - 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 ] &#8658; 2 n a + n d [ 6 n - 1 ] = 1000 2 a + 6 n - 1 d = 1000 n Now, S 6 n = 6 n 2 2 a + 6 n - 1 d = 3 n &#183; 1000 n = 3000

Let $S_1$ be the sum of first $2n$ terms of an arithmetic progression. Let $S_2$ be the sum of first $4n$…

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