Let S 1 be the sum of first 2 n terms of an arithmetic progression. Let S 2 be the sum of first 4 n terms of the same arithmetic progression. If S 2 - S 1 is 1000 , then the sum of the first 6 n terms of the arithmetic progression is equal to:

Step-by-step solution

S 2 n = 2 n 2 2 a + 2 n - 1 d ,   S 4 n = 4 n 2 2 a + 4 n - 1 d ⇒ 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 ] ⇒ 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 · 1000 n = 3000