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 first n terms of an A.P.If S_(2n)=3S_(n), then find the ratio S_(3n)/S_(n)

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

(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

Let S_(n) denote the sum of the first n tem of an A.P.If S_(2n)=3S_(n) then prove that (S_(3n))/(S_(n))=6