If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to

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Here, we are given an A.P. whose sum of n terms is Sn and S_(2m) = 3S_n. We need to find (S_(3m))/(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_(3m) = (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] ................(3) Now, S_(2n) = 3S_n` So, using (2) and (3), we get, `(2n)/2 ( 2a + 2nd - d ) = 3 [n/2 (2a + nd - d)] (2n)/2 (2a + 2nd - d) = (3n)/2 (2a + nd - d)` On further solving, we get, 2(2a + 2nd - d ) = 3 (2a + nd - d) 4a + 4nd - 2d = 6a + 3nd - 3d 2a = nd + d .....................(4) So, `(S_(3n))/(S_n) = ((3n)/2 [ 2a + 3nd -d])/(n/((2)) [ 2a + nd - d ]) Taking n/2 common, we get, S_(3n)/(S_n) = (3(2a + 3nd - d))/(2a + nd - d) =(3(nd + d + 3nd - d))/((nd + d + nd - d)) (Using 4) = (3(4nd))/(2nd) = 6 Therefore, S_(3n)/(S_n) = 6`

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