If `S_1` is the sum of an AP of 'n' odd number of terms and `S_2` be the sum of the terms of series in odd places of the same AP then `S_1/S_2` =

If S_1 is the sum of an AP of n odd number of terms and S_2 be the sum of terms of series in odd places of the same AP, then (S_1)/(S_2)= a.     (2n)/(n+1) b.     n/(n+1) c.     (n+1)/(2n) d.     (n+1)/n

If S_1 is the sum of first 2n terms and S_2 is the sum of first 4n term and S_2-S_1 is equal to 1000 then S_3 sum of first 6n terms of same A.P

If S_1 be the sum of (2n+1) term of an A.P. and S_2 be the sum of its odd terms then prove that S_1: S_2=(2n+1):(n+1)dot

Let S_(1) be the sum of the first n terms of the A.P 8,12,16,... and let S_(2) be the sum of the first n terms of the A.P 17,19,21,... assume n!=0 then S_(1)=S_(2) for

A G.P consists of 2n terms . If the sum of the terms . If the sum of the terms occupying the odd place is S_(1), and that of the terms in the even places is S_(2) then find the common ratio of the progression

If S_n , be the sum of n terms of an A.P ; the value of S_n-2S_(n-1)+ S_(n-2) , is

If S_(n), be the sum of n terms of an A.P; the value of S_(n)-2S_(n-1)+S_(n-2), is

If S_(1), S_(2), S_(3) be the sum of 10, 20, 30 terms respectively of an A.P. Then