Let S_i be the sum of the first n terms of the i -th arithmetic progression. There are 8 such arithmetic progressions. For i = 1, 2, , 8 , the first term of the i -th arithmetic progression is i and its common difference is 2i-1 . If _ i=1 ^8 S_i = 3240 , then the value of n is

Step-by-step solution

The sum of the first n terms of the i -th arithmetic progression is given by S_i = n 2 [2a + (n-1)d] Substituting a = i and d = 2i-1 , we get S_i = n 2 [2i + (n-1)(2i-1)] S_i = n 2 [2i + 2ni - n - 2i + 1] S_i = n 2 [2ni - n + 1] Now, summing S_i from i=1 to 8 : _ i=1 ^8 S_i = _ i=1 ^8 n 2 [2ni - n + 1] _ i=1 ^8 S_i = n 2 [ 2n _ i=1 ^8 i - (n-1) _ i=1 ^8 1 ] _ i=1 ^8 S_i = n 2 [ 2n ( 8 9 2 ) - 8(n-1) ] _ i=1 ^8 S_i = n 2 [72n - 8n + 8] = n 2 [64n + 8] = 32n^2 + 4n Given that _ i=1 ^8 S_i = 3240 , we have: 32n^2 + 4n