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

(1) .Let S_n denote the sum of the first 'n' terms and S_(2n)= 3S_n. Then, the ratio S_(3n):S_n is: (2) let S_1(n) be the sum of the first n terms arithmetic progression 8, 12, 16,..... and S_2(n) be the sum of the first n terms of arithmatic progression 17, 19, 21,....... if S_1(n)=S_2(n) then this common sum is

Find the sum S_n of the cubes of the first n terms of an A.P. and show that the sum of first n terms of the A.P. is a factor of S_n .

If the n^(t h) term of an A.P. is (2n+1), find the sum of first n terms of the A.P.

Let S_n denote the sum of first n terms of an A.P. and S_2n = 3S_n then ratio of S_3n : S_n

Let S_n denotes the sum of the first of n terms of A.P. and S_(2n)=3S_n . then the ratio S_(3n):S_n is equal to

Let S_(n) denote the sum of the first n terms of an A.P.. If S_(4)=16 and S_(6)=-48 , then S_(10) is equal to :

Let S_(n) denote the sum of the first n terms of an A.P.. If S_(4)=16 and S_(6)=-48, then S_(10) is equal to :

The n^(th) term of an A.P is p and the sum of the first n terms is s. The first term is

Let S_(n) denote the sum of the first n terms of an A.P. If S_(2n) = 3S_(n) , then S_(3n) : S_(n) is equal to :

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