If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then S_1/S_2 =

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In the given problem, we are given S1 as the sum of an A.P of ‘n’ odd number of terms and S2 the sum of the terms of the series in odd places. We need to find (S_1)/(S_2) Now, let a1, a2…. an be the n terms of A.P Where n is odd Let d be the common difference of the A.P Then, `S_1 = n /2 [ 2a_1 + ( n - 1) d]` ............(1) And S2 be the sum of the terms of the places in odd places, Where, number of terms = ( n + 1) /2 Common difference = 2d So, `S_2 = ((n + 1)/2 )/2 [2a_1 + ((n+1)/2 - 1) 2d] S_2 = ( n+1)/4 [2a_1 + ((n-1)/2)2d] S_2 = ( n +1)/4 [ 2a _1 + (n-1)d ] .............(2) Now, (S_1)/(S_2) = (n/2[2a_1 + (n-1)d])/((n+1)/4[2a_1 + (n-1)d]) (S_1)/(S_2) = (4n)/(2(n +1)) (S_1)/(S_2) = (2n)/(n + 1) Thus, (S_1)/(S_2) = (2n)/(n + 1)`

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