If Sr denotes the sum of the first r terms of an A.P. Then , S3n: (S2n − Sn) is

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Here, we are given an A.P. whose sum of r terms is Sr. We need to find (S_(3n))/(S_(2n) - S_n). Here we use the following formula for the sum of n terms of an A.P. S_n = n/2 [ 2a + (n -1 ) d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms So, first we find S3n, `S_(3n) = (3n)/2 [ 2a + (3n - 1)d] =(3n)/2 [2a + 3nd - d ]` .................(1) Similarly, `S_(2n) = (2n)/2 [ 2a + (2n - 1 ) d ] = (2n)/2 [2a + 2nd -d]` .................(2) Also, S_n = n/2 [ 2a + (n-1) d] =n/2 [2a + nd - d ] So, using (1), (2) and (3), we get, `(S_(3n))/(S_(2n) - S_n) = ((3n)/2 [2a + 3nd - d])/((2n)/2 [ 2a + 2nd - d ] - n/2 [ 2a + nd - d ]) Taking n/2 common, we get, (S_(3n))/(S_(2n) - S_n) =(3[2a + 3nd - d])/(2[2a + 2nd - d ]- [2a + nd - d]) =(3[2a + 3nd - d])/(4a + 4nd - 2d - 2a - nd + d) =(3[2a + 3nd - d])/(2a + 3nd - d) = 3 Therefore, (S_(3n))/(S_(2n)- S_n )= 3 `

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